Count occurrences of a substring

From Rosetta Code
Task
Count occurrences of a substring
You are encouraged to solve this task according to the task description, using any language you may know.
Task

Create a function,   or show a built-in function,   to count the number of non-overlapping occurrences of a substring inside a string.

The function should take two arguments:

  •   the first argument being the string to search,   and
  •   the second a substring to be searched for.


It should return an integer count.

print countSubstring("the three truths","th")
3

// do not count substrings that overlap with previously-counted substrings:
print countSubstring("ababababab","abab")
2

The matching should yield the highest number of non-overlapping matches.

In general, this essentially means matching from left-to-right or right-to-left   (see proof on talk page).


Other tasks related to string operations:
Metrics
Counting
Remove/replace
Anagrams/Derangements/shuffling
Find/Search/Determine
Formatting
Song lyrics/poems/Mad Libs/phrases
Tokenize
Sequences



11l

print(‘the three truths’.count(‘th’))
print(‘ababababab’.count(‘abab’))
Output:
3
2

360 Assembly

The program uses two ASSIST macros (XDECO,XPRNT) to keep the code as short as possible.

*        Count occurrences of a substring  05/07/2016
COUNTSTR CSECT
         USING  COUNTSTR,R13       base register
         B      72(R15)            skip savearea
         DC     17F'0'             savearea
         STM    R14,R12,12(R13)    prolog
         ST     R13,4(R15)         "
         ST     R15,8(R13)         " 
         LR     R13,R15            "
         MVC    HAYSTACK,=CL32'the three truths'
         MVC    LENH,=F'17'        lh=17
         MVC    NEEDLE,=CL8'th'    needle='th'
         MVC    LENN,=F'2'         ln=2
         BAL    R14,SHOW           call show
         MVC    HAYSTACK,=CL32'ababababab'
         MVC    LENH,=F'11'        lh=11
         MVC    NEEDLE,=CL8'abab'  needle='abab'
         MVC    LENN,=F'4'         ln=4
         BAL    R14,SHOW           call show
         L      R13,4(0,R13)       epilog 
         LM     R14,R12,12(R13)    "
         XR     R15,R15            "
         BR     R14                exit
HAYSTACK DS     CL32               haystack
NEEDLE   DS     CL8                needle
LENH     DS     F                  length(haystack)
LENN     DS     F                  length(needle)
*------- ----   show---------------------------------------------------
SHOW     ST     R14,SAVESHOW       save return address
         BAL    R14,COUNT          count(haystack,needle)
         LR     R11,R0             ic=count(haystack,needle)
         MVC    PG(20),HAYSTACK    output haystack
         MVC    PG+20(5),NEEDLE    output needle
         XDECO  R11,PG+25          output ic
         XPRNT  PG,80              print buffer
         L      R14,SAVESHOW       restore return address
         BR     R14                return to caller
SAVESHOW DS     A                  return address of caller
PG       DC     CL80' '            buffer
*------- ----   count--------------------------------------------------
COUNT    ST     R14,SAVECOUN       save return address
         SR     R7,R7              n=0
         LA     R6,1               istart=1
         L      R10,LENH           lh
         S      R10,LENN           ln
         LA     R10,1(R10)         lh-ln+1
LOOPI    CR     R6,R10             do istart=1 to lh-ln+1
         BH     ELOOPI
         LA     R8,NEEDLE          @needle
         L      R9,LENN            ln
         LA     R4,HAYSTACK-1      @haystack[0]
         AR     R4,R6              +istart
         LR     R5,R9              ln
         CLCL   R4,R8              if substr(haystack,istart,ln)=needle
         BNE    NOTEQ
         LA     R7,1(R7)           n=n+1
         A      R6,LENN            istart=istart+ln
NOTEQ    LA     R6,1(R6)           istart=istart+1
         B      LOOPI
ELOOPI   LR     R0,R7              return(n)
         L      R14,SAVECOUN       restore return address
         BR     R14                return to caller
SAVECOUN DS     A                  return address of caller
*        ----   -------------------------------------------------------
         YREGS
         END    COUNTSTR
Output:
the three truths    th              3
ababababab          abab            2

8080 Assembly

The routine subcnt takes the string pointer in HL and the substring pointer in BC, and returns a 16-bit count in DE.

	org	100h
	jmp	demo
	;;;	Count non-overlapping substrings (BC) in string (HL)
	;;;	Returns amount of matches in DE
subcnt:	lxi	d,0	; Amount of matches
s_scan:	mov	a,m	; Get current character
	ana	a	; End of string?
	rz		; Then stop
	push	b	; Keep start of substring search
	push	h 	; Keep current location in string
s_cmp:	ldax	b	; Get character from substring
	cmp	m	; Compare to curent charracter of search string
	inx	b	; Advance pointers
	inx	h
	jz	s_cmp	; Keep going if they were equal
	ana	a	; Did we reach the end of the substring?
	jz	s_find	; If so, we found a match
	pop	h	; Otherwise, no match - restore search position
	pop	b	; Restore start of substring
	inx	h	; Try next position
	jmp	s_scan
s_find:	inx	d	; We found a match
	pop	b	; Discard start of the search, keep going after match
	pop 	b	; Restore start of substring
	dcx	h	; The comparison routine overshoots by one
	jmp	s_scan
	;;;	Test on a few strings
demo:	lxi	h,pairs
loop:	mov	e,m	; Load string pointer
	inx	h
	mov	d,m
	inx	h
	mov	a,d	; If 0, stop
	ora	e
	rz
	mov	c,m	; Load substring pointer
	inx	h
	mov	b,m
	inx	h
	push	h	; Save example pointer
	xchg		; Put string pointer in HL
	call	subcnt	; Count substrings
	mvi	a,'0'	; Assuming output is <10, print output
	add	e	; (This is true for all examples, and a proper numeric
	mov	e,a	; output routine is big and not relevant.)
	mvi	c,2	; CP/M character output
	call	5
	pop	h	; Restore example pointer
	jmp	loop
pairs:	dw	str1,sub1,str2,sub2,str3,sub3,0
str1:	db	'the three truths',0
sub1:	db	'th',0		; result should be 3
str2:	db	'ababababab',0
sub2:	db	'abab',0	; result should be 2
str3:	db	'cat',0
sub3:	db	'dog',0		; result should be 0
Output:
320

8086 Assembly

	cpu	8086
	org	100h
section	.text
	jmp	demo
	;;;	Count non-overlapping substrings [ES:DI] in [DS:SI]
	;;;	Return count in AX
subcnt:	xor	ax,ax		; Set count to 0
	xor	dl,dl		; Zero to compare to
	mov	bp,di		; Keep copy of substring pointer
.scan:	cmp	dl,[si]		; End of string?
	je	.out		; Then we're done
	mov	bx,si		; Keep copy of search position
	mov	di,bp		; Start at beginning of substring
.cmp:	xor	cx,cx	
	dec	cx
	repe	cmpsb		; Scan until no match 
	dec	si		; Go to first non-match
	dec	di
	cmp	dl,[es:di]	; Reached end of substring?
	je	.match		; Then we found a match
	mov	si,bx 		; If not, continue searching one
	inc	si		; position further
	jmp	.scan
.match:	inc	ax		; Found a match - increment count
	jmp	.scan
.out:	ret
	;;;	Test the routine on a few examples
demo:	mov	si,pairs
.loop:	lodsw			; Load string pointer
	test	ax,ax		; If 0, stop
	jz	.out
	xchg	dx,ax
	lodsw			; Load substring pointer
	xchg	di,ax
	push	si		; Keep example pointer
	xchg	si,dx
	call	subcnt		; Count substrings
	call	prax		; Print amount of substrings
	pop	si		; Restore example pointer
	jmp	.loop
.out:	ret
	;;;	Print AX as number
prax:	mov	bx,num		; Pointer to end of number string
	mov	cx,10 		; Divisor
.dgt:	xor	dx,dx		; Divide by 10 
	div	cx
	add	dl,'0'		; Add ASCII 0 to remainder
	dec	bx		; Store digit
	mov	[bx],dl
	test	ax,ax		; If number not zero yet
	jnz	.dgt		; Find rest of digits
	mov	dx,bx		; Print number string
	mov	ah,9
	int	21h
	ret
section	.data
	db	'*****'		; Output number placeholder
num:	db	' $'
	;;;	Examples
pairs:	dw	.str1,.sub1,.str2,.sub2,.str3,.sub3,0
.str1:	db	'the three truths',0
.sub1:	db	'th',0		; result should be 3
.str2:	db	'ababababab',0
.sub2:	db	'abab',0	; result should be 2
.str3:	db	'cat',0
.sub3:	db	'dog',0		; result should be 0
Output:
3 2 0

AArch64 Assembly

Works with: as version Raspberry Pi 3B version Buster 64 bits
or android 64 bits with application Termux
/* ARM assembly AARCH64 Raspberry PI 3B */
/*  program strcptsub64.s   */

/************************************/
/* Constantes                       */
/************************************/
/* for this file see task include a file in language AArch64 assembly*/
.include "../includeConstantesARM64.inc" 

/************************************/
/* Initialized data                 */
/************************************/
.data
szMessResult:         .asciz "Result: "
szString:             .asciz "the three truths" 
szSubString:          .asciz "th"
szString1:             .asciz "ababababab" 
szSubString1:          .asciz "abab"
szCarriageReturn:     .asciz "\n"
szMessStart:          .asciz "Program 64 bits start.\n"
/************************************/
/* UnInitialized data               */
/************************************/
.bss 
sZoneConv:            .skip 24
/************************************/
/*  code section                    */
/************************************/
.text
.global main   
main:                      // entry of program
    ldr x0,qAdrszMessStart
    bl affichageMess
    ldr x0,qAdrszString
    ldr x1,qAdrszSubString
    bl countSubString
    ldr x0,qAdrszString1
    ldr x1,qAdrszSubString1
    bl countSubString

100:                       // standard end of the program
    mov x0, #0             // return code
    mov x8, #EXIT          // request to exit program
    svc 0                  // perform the system call
qAdrszString:             .quad  szString
qAdrszSubString:          .quad  szSubString
qAdrszString1:            .quad  szString1
qAdrszSubString1:         .quad  szSubString1
qAdrsZoneConv:            .quad  sZoneConv
qAdrszMessResult:         .quad  szMessResult
qAdrszCarriageReturn:     .quad  szCarriageReturn
qAdrszMessStart:          .quad  szMessStart
/***************************************************/
/*   count sub string of string                    */
/***************************************************/
/* r0 contains a string */
/* r1 contains a substring */
/* r0 return substring count */
countSubString:
    stp x1,lr,[sp,-16]!
    stp x2,x3,[sp,-16]! 
    stp x4,x5,[sp,-16]! 
    stp x6,x7,[sp,-16]! 
    mov x4,#0            // counter
    mov x3,#0            // index string
    mov x5,#0            // index substring
1:
    ldrb w6,[x0,x5]      // load byte of string
    ldrb w7,[x1,x3]      // load byte of substring
    cmp x6,x7            // compare byte
    bne 2f               // not equal
    cmp x6,#0            // string end ?
    beq 3f               // yes 
    add x5,x5,#1         // else increment index
    add x3,x3,#1
    b 1b                 // and loop
2:                       // characters not equal
    cmp x6,#0            // end string ?
    beq 4f
    cmp x7,#0            // end substring ?
    add x6,x4,1
    csel x4,x6,x4,eq     // yes -> increment counter 
    mov x3,#0            // raz index substring
    add x5,x5,#1         // increment string index
    b 1b                 // and loop
3:                       // end string and end substring
    add x4,x4,#1         // increment counter
4:                       // result display
    mov x0,x4
    ldr x1,qAdrsZoneConv
    bl conversion10
    ldr x0,qAdrszMessResult
    bl affichageMess
    ldr x0,qAdrsZoneConv
    bl affichageMess
    ldr x0,qAdrszCarriageReturn
    bl affichageMess
    
    mov x0,x4
 100:
    ldp x6,x7,[sp],16
    ldp x4,x5,[sp],16
    ldp x2,x3,[sp],16
    ldp x1,lr,[sp],16
    ret 
/***************************************************/
/*      ROUTINES INCLUDE                           */
/***************************************************/
/* for this file see task include a file in language AArch64 assembly*/
.include "../includeARM64.inc"
Output:
Program 64 bits start.
Result: 3
Result: 2

Action!

BYTE FUNC CountSubstring(CHAR ARRAY s,sub)
  BYTE i,j,res,found

  i=1 res=0
  WHILE i-1+sub(0)<=s(0)
  DO
    found=1
    FOR j=1 TO sub(0)
    DO
      IF s(j+i-1)#sub(j) THEN
        found=0
        EXIT
      FI
    OD

    IF found=1 THEN
      i==+sub(0)
      res==+1
    ELSE
      i==+1
    FI
  OD
RETURN (res)

PROC Test(CHAR ARRAY s,sub)
  BYTE c

  c=CountSubstring(s,sub)
  PrintF("%B ""%S"" in ""%S""%E",c,sub,s)
RETURN

PROC Main()
  Test("the three truths","th")
  Test("ababababab","abab")
  Test("11111111","11")
  Test("abcdefg","123")
RETURN
Output:

Screenshot from Atari 8-bit computer

3 "th" in "the three truths"
2 "abab" in "ababababab"
4 "11" in "11111111"
0 "123" in "abcdefg"

Ada

with Ada.Strings.Fixed, Ada.Integer_Text_IO;

procedure Substrings is
begin
   Ada.Integer_Text_IO.Put (Ada.Strings.Fixed.Count (Source  => "the three truths",
                                                     Pattern => "th"));
   Ada.Integer_Text_IO.Put (Ada.Strings.Fixed.Count (Source  => "ababababab",
                                                     Pattern => "abab"));
end Substrings;
Output:
          3          2

ALGOL 68

Works with: ALGOL 68 version Revision 1 - no extensions to language used.
Works with: ALGOL 68G version Any - tested with release 1.18.0-9h.tiny.

Algol68 has no build in function to do this task, hence the next to create a count string in string routine.

#!/usr/local/bin/a68g --script #

PROC count string in string = (STRING needle, haystack)INT: (
  INT start:=LWB haystack, next, out:=0;
  FOR count WHILE string in string(needle, next, haystack[start:]) DO
    start+:=next+UPB needle-LWB needle;
    out:=count
  OD;
  out
);

printf(($d" "$,
  count string in string("th", "the three truths"),    # expect 3 #
  count string in string("abab", "ababababab"),        # expect 2 #
  count string in string("a*b", "abaabba*bbaba*bbab"), # expect 2 #
  $l$
))
3 2 2

Apex

Apex example for 'Count occurrences of a substring'.

String substr = 'ABC';
String str = 'ABCZZZABCYABCABCXXABC';
Integer substrLen = substr.length();
Integer count = 0;
Integer index = str.indexOf(substr);
while (index >= 0) {
    count++;
    str = str.substring(index+substrLen);
    index = str.indexOf(substr);
}
System.debug('Count String : '+count);
Count String : 5

APL

Works with: Dyalog APL
csubs{0=x⊃⍸⍵:0  1+⍺∇(¯1+x+⍴)}
Output:
      'th' 'abab' 'dog' csubs¨ 'the three truths' 'ababababab' 'cat'
3 2 0

AppleScript

This is a good example of the kind of problem to which standard libraries (a regex library in this case) would offer most languages a simple and immediate solution. AppleScript, however, for want of various basics like regex and math library functions, can require scripters to draw on the supplementary resources of Bash, using the built-in do shell script function.

A slightly faster approach to seeking outside library help has, however, become possible since OS X 10.10 added JavaScript as an osalang sibling for AppleScript. Rather than using the resources of Bash, we can more quickly use AppleScript's built in ObjC interface to pass a subproblem over to the more richly endowed JavaScript core of JavaScript for Automation.

Here we use a generic evalOSA(language, code) function to apply a JavaScript for Automation regex to a pair of AppleScript strings, using OSAKit.

use framework "OSAKit"

on run
    {countSubstring("the three truths", "th"), ¬
        countSubstring("ababababab", "abab")}
end run

on countSubstring(str, subStr)
    return evalOSA("JavaScript", "var matches = '" & str & "'" & ¬
        ".match(new RegExp('" & subStr & "', 'g'));" & ¬
        "matches ? matches.length : 0") as integer
end countSubstring

-- evalOSA :: ("JavaScript" | "AppleScript") -> String -> String
on evalOSA(strLang, strCode)
    
    set ca to current application
    set oScript to ca's OSAScript's alloc's initWithSource:strCode ¬
        |language|:(ca's OSALanguage's languageForName:(strLang))
    
    set {blnCompiled, oError} to oScript's compileAndReturnError:(reference)
    
    if blnCompiled then
        set {oDesc, oError} to oScript's executeAndReturnError:(reference)
        if (oError is missing value) then return oDesc's stringValue as text
    end if
    
    return oError's NSLocalizedDescription as text
end evalOSA
Output:
{3, 2}

The above assertions notwithstanding, it's always been possible to use AppleScript's text item delimiters for this purpose with its native strings, except that the TIDs have only observed the current considering/ignoring state with the 'unicode text' class, which was introduced around Mac OS X 10.4 and became AppleScript's only native text class with the introduction of AppleScript 2.0 in Mac OS X 10.5.

on countSubstring(theString, theSubstring)
    set astid to AppleScript's text item delimiters
    set AppleScript's text item delimiters to theSubstring
    set substringCount to (count theString's text items) - 1
    set AppleScript's text item delimiters to astid
    
    return substringCount
end countSubstring

{countSubstring("the three truths", "th"), countSubstring("ababababab", "abab")}
Output:
{3, 2}

ARM Assembly

Works with: as version Raspberry Pi
or android 32 bits with application Termux
/* ARM assembly Raspberry PI  */
/*  program strcptsub.s   */

/************************************/
/* Constantes                       */
/************************************/
/* for this file see task include a file in language ARM assembly*/
.include "../constantes.inc"

/************************************/
/* Initialized data                 */
/************************************/
.data
szMessResult:         .asciz "Result: "
szString:             .asciz "the three truths" 
szSubString:          .asciz "th"
szString1:             .asciz "ababababab" 
szSubString1:          .asciz "abab"
szCarriageReturn:     .asciz "\n"
szMessStart:          .asciz "Program 32 bits start.\n"
/************************************/
/* UnInitialized data               */
/************************************/
.bss 
sZoneConv:            .skip 24
/************************************/
/*  code section                    */
/************************************/
.text
.global main   
main:                      @ entry of program
    ldr r0,iAdrszMessStart
    bl affichageMess
    ldr r0,iAdrszString
    ldr r1,iAdrszSubString
    bl countSubString
    ldr r0,iAdrszString1
    ldr r1,iAdrszSubString1
    bl countSubString

100:                       @ standard end of the program
    mov r0, #0             @ return code
    mov r7, #EXIT          @ request to exit program
    svc 0                  @ perform the system call
iAdrszString:             .int szString
iAdrszSubString:          .int szSubString
iAdrszString1:            .int szString1
iAdrszSubString1:         .int szSubString1
iAdrsZoneConv:            .int sZoneConv
iAdrszMessResult:         .int szMessResult
iAdrszCarriageReturn:     .int szCarriageReturn
iAdrszMessStart:          .int szMessStart
/***************************************************/
/*   count sub string of string                    */
/***************************************************/
/* r0 contains a string */
/* r1 contains a substring */
/* r0 return substring count */
countSubString:
    push {r1-r7,lr}      @ save registers
    mov r4,#0            @ counter
    mov r3,#0            @ index string
    Mov r5,#0            @ index substring
1:
    ldrb r6,[r0,r5]      @ load byte of string
    ldrb r7,[r1,r3]      @ load byte of substring
    cmp r6,r7            @ compare byte
    bne 2f               @ not equal
    cmp r6,#0            @ string end ?
    beq 3f               @ yes 
    add r5,r5,#1         @ else increment index
    add r3,r3,#1
    b 1b                 @ and loop
2:                       @ characters not equal
    cmp r6,#0            @ end string ?
    beq 4f
    cmp r7,#0            @ end substring ?
    addeq r4,r4,#1       @ yes -> increment counter 
    mov r3,#0            @ raz index substring
    add r5,r5,#1         @ increment string index
    b 1b                 @ and loop
3:                       @ end string and end substring
    add r4,r4,#1         @ increment counter
4:                       @ result display
    mov r0,r4
    ldr r1,iAdrsZoneConv
    bl conversion10
    ldr r0,iAdrszMessResult
    bl affichageMess
    ldr r0,iAdrsZoneConv
    bl affichageMess
    ldr r0,iAdrszCarriageReturn
    bl affichageMess
    
    mov r0,r4
 100:
    pop {r1-r7,pc}  
/***************************************************/
/*      ROUTINES INCLUDE                           */
/***************************************************/
/* for this file see task include a file in language ARM assembly*/
.include "../affichage.inc"
Output:
Program 32 bits start.
Result: 3
Result: 2

Arturo

countOccurrences: function [str, substr]-> size match str substr

loop [["the three truths" "th"] ["ababababab" "abab"]] 'pair ->
    print [
        ~"occurrences of '|last pair|' in '|first pair|':" 
        countOccurrences first pair last pair
    ]
Output:
occurrences of 'th' in 'the three truths': 3 
occurrences of 'abab' in 'ababababab': 2

AutoHotkey

While it is simple enough to parse the string, AutoHotkey has a rather unconventional method which outperforms this. StringReplace sets the number of replaced strings to ErrorLevel.

MsgBox % countSubstring("the three truths","th") ; 3
MsgBox % countSubstring("ababababab","abab")     ; 2

CountSubstring(fullstring, substring){
   StringReplace, junk, fullstring, %substring%, , UseErrorLevel
   return errorlevel
}

AWK

#
# countsubstring(string, pattern)
#   Returns number of occurrences of pattern in string
#   Pattern treated as a literal string (regex characters not expanded)
#
function countsubstring(str, pat,    len, i, c) {
  c = 0
  if( ! (len = length(pat) ) ) 
    return 0
  while(i = index(str, pat)) {
    str = substr(str, i + len)
    c++
  }
  return c
}
#
# countsubstring_regex(string, regex_pattern)
#   Returns number of occurrences of pattern in string
#   Pattern treated as regex       
#
function countsubstring_regex(str, pat,    c) {
  c = 0
  c += gsub(pat, "", str)
  return c
}
BEGIN {
  print countsubstring("[do&d~run?d!run&>run&]", "run&")
  print countsubstring_regex("[do&d~run?d!run&>run&]", "run[&]")
  print countsubstring("the three truths","th")
}
Output:
$ awk -f countsubstring.awk
2
2
3

BaCon

FUNCTION Uniq_Tally(text$, part$)
    LOCAL x
    WHILE TALLY(text$, part$)
        INCR x
        text$ = MID$(text$, INSTR(text$, part$)+LEN(part$))
    WEND
    RETURN x
END FUNCTION

PRINT "the three truths - th: ", Uniq_Tally("the three truths", "th")
PRINT "ababababab - abab: ", Uniq_Tally("ababababab", "abab")
Output:
the three truths - th: 3
ababababab - abab: 2

BASIC

Works with: QBasic

In FreeBASIC, this needs to be compiled with -lang qb or -lang fblite.

DECLARE FUNCTION countSubstring& (where AS STRING, what AS STRING)

PRINT "the three truths, th:", countSubstring&("the three truths", "th")
PRINT "ababababab, abab:", countSubstring&("ababababab", "abab")

FUNCTION countSubstring& (where AS STRING, what AS STRING)
    DIM c AS LONG, s AS LONG
    s = 1 - LEN(what)
    DO
        s = INSTR(s + LEN(what), where, what)
        IF 0 = s THEN EXIT DO
        c = c + 1
    LOOP
    countSubstring = c
END FUNCTION
Output:
the three truths, th:        3
ababababab, abab:            2

See also: Liberty BASIC, PowerBASIC, PureBasic.

Applesoft BASIC

10 F$ = "TH"
20 S$ = "THE THREE TRUTHS"
30 GOSUB 100"COUNT SUBSTRING
40 PRINT R
50 F$ = "ABAB"
60 S$ = "ABABABABAB"
70 GOSUB 100"COUNT SUBSTRING
80 PRINT R
90 END

100 R = 0
110 F = LEN(F$)
120 S = LEN(S$)
130 IF F > S THEN RETURN
140 IF F = 0 THEN RETURN
150 IF F = S AND F$ = S$ THEN R = 1 : RETURN
160 FOR I = 1 TO S - F
170     IF F$ = MID$(S$, I, F) THEN R = R + 1 : I = I + F - 1
180 NEXT I
190 RETURN

BASIC256

Translation of: Run BASIC
print countSubstring("the three truths","th")
print countSubstring("ababababab","abab")
end

function countSubstring(s$,find$)
	i = 1
	while instr(s$,find$,i) <> 0
		countSubstring += 1
		i = instr(s$,find$,i) + length(find$)
	end while
end function
Output:
Igual que la entrada de Run BASIC.

Chipmunk Basic

Works with: Chipmunk Basic version 3.6.4
Works with: Applesoft BASIC
Works with: MSX_BASIC
Works with: PC-BASIC version any
Works with: QBasic
Translation of: Applesoft BASIC
10 CLS : REM  10 HOME for Applesoft BASIC
20 F$ = "TH"
30 S$ = "THE THREE TRUTHS"
40 GOSUB 110: REM COUNT SUBSTRING
50 PRINT R
60 F$ = "ABAB"
70 S$ = "ABABABABAB"
80 GOSUB 110: REM COUNT SUBSTRING
90 PRINT R
100 END
110 R = 0
120 F = LEN(F$)
130 S = LEN(S$)
140 IF F > S THEN RETURN
150 IF F = 0 THEN RETURN
160 IF F = S AND F$ = S$ THEN R = 1: RETURN
170 FOR I = 1 TO S - F
180   IF F$ = MID$(S$, I, F) THEN R = R + 1: I = I + F - 1
190 NEXT I
200 RETURN

GW-BASIC

The Chipmunk Basic solution works without any changes.

IS-BASIC

100 INPUT PROMPT "String:    ":TXT$
110 INPUT PROMPT "Substring: ":SUB$
120 PRINT COUNT(LCASE$(TXT$),LCASE$(SUB$))
130 DEF COUNT(TXT$,SUB$)
140   LET N=0:LET PO=1
150   DO
160     LET PO=POS(TXT$,SUB$,PO)
170     IF PO THEN LET N=N+1:LET PO=PO+LEN(SUB$)
180   LOOP UNTIL PO=0
190   LET COUNT=N
200 END DEF

MSX Basic

The Chipmunk Basic solution works without any changes.

Sinclair ZX81 BASIC

Works with 1k of RAM.

 10 LET S$="THE THREE TRUTHS"
 20 LET U$="TH"
 30 GOSUB 100
 40 PRINT N
 50 LET S$="ABABABABAB"
 60 LET U$="ABAB"
 70 GOSUB 100
 80 PRINT N
 90 STOP
100 LET N=0
110 LET I=0
120 LET I=I+1
130 IF I+LEN U$>LEN S$ THEN RETURN
140 IF S$(I TO I+LEN U$-1)<>U$ THEN GOTO 120
150 LET N=N+1
160 LET I=I+LEN U$
170 GOTO 130

True BASIC

Translation of: QBasic
FUNCTION countsubstring(where$, what$)
    LET c = 0
    LET s = 1-LEN(what$)
    DO
       LET s = POS(where$,what$,s+LEN(what$))
       IF 0 = s THEN EXIT DO
       LET c = c+1
    LOOP
    LET countsubstring = c
END FUNCTION

PRINT "the three truths, th:", countSubstring("the three truths", "th")
PRINT "ababababab, abab:", countSubstring("ababababab", "abab")
END

Yabasic

Translation of: Run BASIC
print countSubstring("the three truths","th")
print countSubstring("ababababab","abab")
end

sub countSubstring(s$,find$)
	countSubstring = 0
	i = 1
	while instr(s$,find$,i) <> 0
		countSubstring = countSubstring + 1
		i = instr(s$,find$,i) + len(find$)
	end while
	return countSubstring
end sub
Output:
Igual que la entrada de Run BASIC.

Batch File

@echo off
setlocal enabledelayedexpansion

	::Main
call :countString "the three truths","th"
call :countString "ababababab","abab"
pause>nul
exit /b
	::/Main

	::Procedure
:countString
	set input=%~1
	set cnt=0

	:count_loop
	set trimmed=!input:*%~2=!
	if "!trimmed!"=="!input!" (echo.!cnt!&goto :EOF)
	set input=!trimmed!
	set /a cnt+=1
	goto count_loop
Output:
3
2

BBC BASIC

      tst$ = "the three truths"
      sub$ = "th"
      PRINT ; FNcountSubstring(tst$, sub$) " """ sub$ """ in """ tst$ """"
      tst$ = "ababababab"
      sub$ = "abab"
      PRINT ; FNcountSubstring(tst$, sub$) " """ sub$ """ in """ tst$ """"
      END
      
      DEF FNcountSubstring(A$, B$)
      LOCAL I%, N%
      I% = 1 : N% = 0
      REPEAT
        I% = INSTR(A$, B$, I%)
        IF I% THEN N% += 1 : I% += LEN(B$)
      UNTIL I% = 0
      = N%
Output:
3 "th" in "the three truths"
2 "abab" in "ababababab"

BCPL

get "libhdr"

let countsubstr(str, match) = valof
$(  let i, count = 1, 0
    while i <= str%0 do
        test valof
        $(  for j = 1 to match%0
                unless match%j = str%(i+j-1)
                    resultis false
            resultis true
        $)
        then
        $(  count := count + 1
            i := i + match%0
        $)
        else
            i := i + 1
    resultis count
$)

let show(str, match) be
    writef("*"%S*" in *"%S*": %N*N", 
        match, str, countsubstr(str, match))

let start() be
$(  show("the three truths", "th")
    show("ababababab", "abab")
    show("cat", "dog")
$)
Output:
"th" in "the three truths": 3
"abab" in "ababababab": 2
"dog" in "cat": 0

BQN

/𝕨⍷𝕩 finds locations of substrings, rest of the function suppresses overlapping substrings.

Find{i/𝕨𝕩, i/˜i≥»0,(𝕨)+⊢`i}

•Show "abab" Find "ababababab" 
•Show "th" Find "the three truths"
2
3

Using strings.bqn from bqn-libs, another solution is Find←+´Locate, since Locate performs a non-overlapping search.

Bracmat

  ( count-substring
  =   n S s p
    .     0:?n:?p
        & !arg:(?S.?s)
        & @( !S
           :   ?
               ( [!p ? !s [?p ?
               & !n+1:?n
               & ~
               )
           )
      | !n
  )
& out$(count-substring$("the three truths".th))
& out$(count-substring$(ababababab.abab))
& ;
Output:
3
2

C

#include <stdio.h>
#include <string.h>

int match(const char *s, const char *p, int overlap)
{
        int c = 0, l = strlen(p);

        while (*s != '\0') {
                if (strncmp(s++, p, l)) continue;
                if (!overlap) s += l - 1;
                c++;
        }
        return c;
}

int main()
{
        printf("%d\n", match("the three truths", "th", 0));
        printf("overlap:%d\n", match("abababababa", "aba", 1));
        printf("not:    %d\n", match("abababababa", "aba", 0));
        return 0;
}

Alternate version:

#include <stdio.h>
#include <string.h>

// returns count of non-overlapping occurrences of 'sub' in 'str'
int countSubstring(const char *str, const char *sub)
{
    int length = strlen(sub);
    if (length == 0) return 0;
    int count = 0;
    for (str = strstr(str, sub); str; str = strstr(str + length, sub))
        ++count;
    return count;
}

int main()
{
    printf("%d\n", countSubstring("the three truths", "th"));
    printf("%d\n", countSubstring("ababababab", "abab"));
    printf("%d\n", countSubstring("abaabba*bbaba*bbab", "a*b"));

    return 0;
}
Output:
3
2
2

C#

using System;

class SubStringTestClass
{
   public static int CountSubStrings(this string testString, string testSubstring)
   {
        int count = 0;
            
        if (testString.Contains(testSubstring))
        {
            for (int i = 0; i < testString.Length; i++)
            {
                if (testString.Substring(i).Length >= testSubstring.Length)
                {
                    bool equals = testString.Substring(i, testSubstring.Length).Equals(testSubstring);
                    if (equals)
                    {
                        count++;
                        i += testSubstring.Length - 1;  // Fix: Don't count overlapping matches
                    }
                }
            }
        }
        return count;
   }
}

Using C# 6.0's expression-bodied member, null-conditional operator, and coalesce operator features:

using System;
class SubStringTestClass
{
   public static int CountSubStrings(this string testString, string testSubstring) =>
       testString?.Split(new [] { testSubstring }, StringSplitOptions.None)?.Length - 1 ?? 0;
}

C++

#include <iostream>
#include <string>

// returns count of non-overlapping occurrences of 'sub' in 'str'
int countSubstring(const std::string& str, const std::string& sub)
{
    if (sub.length() == 0) return 0;
    int count = 0;
    for (size_t offset = str.find(sub); offset != std::string::npos;
	 offset = str.find(sub, offset + sub.length()))
    {
        ++count;
    }
    return count;
}

int main()
{
    std::cout << countSubstring("the three truths", "th")    << '\n';
    std::cout << countSubstring("ababababab", "abab")        << '\n';
    std::cout << countSubstring("abaabba*bbaba*bbab", "a*b") << '\n';

    return 0;
}
Output:
3
2
2

Clojure

Use a sequence of regexp matches to count occurrences.

(defn re-quote
  "Produces a string that can be used to create a Pattern
   that would match the string text as if it were a literal pattern.
   Metacharacters or escape sequences in text will be given no special
   meaning"
  [text]
  (java.util.regex.Pattern/quote text))

(defn count-substring [txt sub]
  (count (re-seq (re-pattern (re-quote sub)) txt)))

Use the trick of blank replacement and maths to count occurrences.

(defn count-substring1 [txt sub]
  (/ (- (count txt) (count (.replaceAll txt sub "")))
     (count sub)))

A Java 8 stream-based solution, which should avoid creation of temporary strings (though it will produce temporary MatchResult instances).

(defn count-substring2 [txt sub]
  (-> sub
    (re-quote)
    (re-pattern)
    (re-matcher txt)
    (.results)
    (.count)))

COBOL

INSPECT can be used for this task without having to create a function.

       IDENTIFICATION DIVISION.
       PROGRAM-ID. testing.

       DATA DIVISION.
       WORKING-STORAGE SECTION.
       01  occurrences             PIC 99.

       PROCEDURE DIVISION.
           INSPECT "the three truths" TALLYING occurrences FOR ALL "th"
           DISPLAY occurrences

           MOVE 0 TO occurrences
           INSPECT "ababababab" TALLYING occurrences FOR ALL "abab"
           DISPLAY occurrences
           
           MOVE 0 TO occurrences
           INSPECT "abaabba*bbaba*bbab" TALLYING occurrences
               FOR ALL "a*b"
           DISPLAY occurrences

           GOBACK
           .
Output:
03
02
02

CoffeeScript

countSubstring = (str, substr) ->
  n = 0
  i = 0
  while (pos = str.indexOf(substr, i)) != -1
    n += 1
    i = pos + substr.length
  n

console.log countSubstring "the three truths", "th"
console.log countSubstring "ababababab", "abab"

Common Lisp

(defun count-sub (str pat)
  (loop with z = 0 with s = 0 while s do
	(when (setf s (search pat str :start2 s))
	  (incf z) (incf s (length pat)))
	finally (return z)))

(count-sub "ababa" "ab")  ; 2
(count-sub "ababa" "aba") ; 1

Cowgol

include "cowgol.coh";

sub countSubstring(str: [uint8], match: [uint8]): (count: uint8) is
    count := 0;
    
    while [str] != 0 loop
        var find := match;
        var loc := str;
        while [loc] == [find] loop
            find := @next find;
            loc := @next loc;
        end loop;
        if [find] == 0 then 
            str := loc;
            count := count + 1;
        else    
            str := @next str;
        end if;
    end loop;
end sub;

print_i8(countSubstring("the three truths","th")); # should print 3
print_nl();
print_i8(countSubstring("ababababab","abab")); # should print 2
print_nl();
print_i8(countSubstring("cat","dog")); # should print 0
print_nl();
Output:
3
2
0

D

void main() {
    import std.stdio, std.algorithm;

    "the three truths".count("th").writeln;
    "ababababab".count("abab").writeln;
}
Output:
3
2

Delphi

program OccurrencesOfASubstring;

{$APPTYPE CONSOLE}

uses StrUtils;

function CountSubstring(const aString, aSubstring: string): Integer;
var
  lPosition: Integer;
begin
  Result := 0;
  lPosition := PosEx(aSubstring, aString);
  while lPosition <> 0 do
  begin
    Inc(Result);
    lPosition := PosEx(aSubstring, aString, lPosition + Length(aSubstring));
  end;
end;

begin
  Writeln(CountSubstring('the three truths', 'th'));
  Writeln(CountSubstring('ababababab', 'abab'));
end.

Draco

proc countSubstring(*char string, substring) word:
    word count;
    *char pos, loc;
    
    count := 0;
    while string* /= '\e' do
        pos := substring;
        loc := string;
        while loc* = pos* do
            loc := loc + 1;
            pos := pos + 1
        od;
        if pos* = '\e' then
            string := loc;
            count := count + 1
        else
            string := string + 1
        fi
    od;
    count
corp

proc main() void:
    writeln(countSubstring("the three truths", "th"));
    writeln(countSubstring("ababababab", "abab"))
corp
Output:
3
2

Dyalect

func countSubstring(str, val) {
    var idx = 0
    var count = 0
    while true {
        idx = str.IndexOf(val, idx)
        if idx == -1 {
            break
        }
        idx += val.Length()
        count += 1
    }
    return count
}

print(countSubstring("the three truths", "th"))
print(countSubstring("ababababab", "abab"))
Output:
3
2

Déjà Vu

!. count "the three truths" "th"
!. count "ababababab" "abab"
Output:
3
2

EasyLang

func count str$ pat$ .
   ind = 1
   while ind + len pat$ - 1 <= len str$
      if substr str$ ind len pat$ = pat$
         cnt += 1
         ind += len pat$
      else
         ind += 1
      .
   .
   return cnt
.
print count "the three truths" "th"
print count "ababababab" "abab"
print count "11111111" "11"
print count "11111111" "12"
print count "12" "12"

EchoLisp

;; from Racket
(define count-substring
   (compose length regexp-match*))

(count-substring "aab" "graabaabdfaabgh") ;; substring
     3
(count-substring "/ .e/" "Longtemps je me suis couché de bonne heure") ;; regexp
     4

EGL

Works with: EDT

The "remove and count the difference" and "manual loop" methods. Implementation includes protection from empty source and search strings.

program CountStrings
	
    function main()
        SysLib.writeStdout("Remove and Count:");
        SysLib.writeStdout(countSubstring("th", "the three truths"));
        SysLib.writeStdout(countSubstring("abab", "ababababab"));
        SysLib.writeStdout(countSubstring("a*b", "abaabba*bbaba*bbab"));
        SysLib.writeStdout(countSubstring("a", "abaabba*bbaba*bbab"));
        SysLib.writeStdout(countSubstring(" ", "abaabba*bbaba*bbab"));
        SysLib.writeStdout(countSubstring("", "abaabba*bbaba*bbab"));
        SysLib.writeStdout(countSubstring("a", ""));
        SysLib.writeStdout(countSubstring("", ""));

        SysLib.writeStdout("Manual Loop:");
        SysLib.writeStdout(countSubstringWithLoop("th", "the three truths"));
        SysLib.writeStdout(countSubstringWithLoop("abab", "ababababab"));
        SysLib.writeStdout(countSubstringWithLoop("a*b", "abaabba*bbaba*bbab"));
        SysLib.writeStdout(countSubstringWithLoop("a", "abaabba*bbaba*bbab"));
        SysLib.writeStdout(countSubstringWithLoop(" ", "abaabba*bbaba*bbab"));
        SysLib.writeStdout(countSubstringWithLoop("", "abaabba*bbaba*bbab"));
        SysLib.writeStdout(countSubstringWithLoop("a", ""));
        SysLib.writeStdout(countSubstringWithLoop("", ""));
    end

    function countSubstring(substr string in, str string in) returns(int)
        if(str.length() > 0 and substr.length() > 0)
	    return (str.length() - str.replaceStr(subStr, "").length()) / subStr.length();
	else
	    return 0;
	end
    end 
	
    function countSubstringWithLoop(substr string in, str string in) returns(int)
        count int = 0;
        loc, index int = 1;
        strlen int = str.length();
        substrlen int = substr.length();
		
        if(strlen > 0 and substrlen > 0)
            while(loc != 0 and index <= strlen)
                loc = str.indexOf(substr, index);
                if(loc > 0)
                    count += 1;
                    index = loc + substrlen;
                end
            end
        end
        return count;
    end

end
Output:
Remove and Count:
3
2
2
7
0
0
0
0
Manual Loop:
3
2
2
7
0
0
0
0

Eiffel

class
	APPLICATION
inherit
	ARGUMENTS
create
	make
feature {NONE} -- Initialization
	make
			-- Run application.
		do
			occurance := 0
			from
				index := 1
			until
				index > text.count
			loop
				temp := text.fuzzy_index(search_for, index, 0)
				if
					temp /= 0
				then
					index := temp + search_for.count
					occurance := occurance + 1
				else
					index := text.count + 1
				end
			end
			print(occurance)
		end

	index:INTEGER
	temp:INTEGER
	occurance:INTEGER
	text:STRING = "ababababab"
	search_for:STRING = "abab"
end

Elixir

countSubstring = fn(_, "") -> 0
                   (str, sub) -> length(String.split(str, sub)) - 1 end

data = [ {"the three truths", "th"},
         {"ababababab", "abab"},
         {"abaabba*bbaba*bbab", "a*b"},
         {"abaabba*bbaba*bbab", "a"},
         {"abaabba*bbaba*bbab", " "},
         {"abaabba*bbaba*bbab", ""},
         {"", "a"},
         {"", ""} ]

Enum.each(data, fn{str, sub} ->
  IO.puts countSubstring.(str, sub)
end)
Output:
3
2
2
7
0
0
0
0

Erlang

%% Count non-overlapping substrings in Erlang for the rosetta code wiki.
%% Implemented by J.W. Luiten

-module(substrings).
-export([main/2]).

%% String and Sub exhausted, count a match and present result
match([], [], _OrigSub, Acc) ->
  Acc+1;

%% String exhausted, present result
match([], _Sub, _OrigSub, Acc) ->
  Acc;
	
%% Sub exhausted, count a match
match(String, [], Sub, Acc) ->
  match(String, Sub, Sub, Acc+1);

%% First character matches, advance
match([X|MainTail], [X|SubTail], Sub, Acc) ->
  match(MainTail, SubTail, Sub, Acc);

%% First characters do not match. Keep scanning for sub in remainder of string
match([_X|MainTail], [_Y|_SubTail], Sub, Acc)->
  match(MainTail, Sub, Sub, Acc).
	
main(String, Sub) ->
   match(String, Sub, Sub, 0).
Command:
substrings:main("ababababab","abab").
Output:
2

Alternative using built in functions:

main( String, Sub ) -> erlang:length( binary:split(binary:list_to_bin(String), binary:list_to_bin(Sub), [global]) ) - 1.

Euphoria

function countSubstring(sequence s, sequence sub)
    integer from,count
    count = 0
    from = 1
    while 1 do
        from = match_from(sub,s,from)
        if not from then
            exit
        end if
        from += length(sub)
        count += 1
    end while
    return count
end function

? countSubstring("the three truths","th")
? countSubstring("ababababab","abab")
Output:
3
2

F#

"Remove and count the difference" method, as shown by J, Java, ...

open System

let countSubstring (where :string) (what : string) =
    match what with
    | "" -> 0 // just a definition; infinity is not an int
    | _ -> (where.Length - where.Replace(what, @"").Length) / what.Length
    

[<EntryPoint>]
let main argv =
    let show where what =
        printfn @"countSubstring(""%s"", ""%s"") = %d" where what (countSubstring where what)
    show "the three truths" "th"
    show "ababababab" "abab"
    show "abc" ""
    0
countSubstring("the three truths", "th") = 3
countSubstring("ababababab", "abab") = 2
countSubstring("abc", "") = 0

Factor

USING: math sequences splitting ;
: occurences ( seq subseq -- n ) split-subseq length 1 - ;

Forth

: str-count ( s1 len s2 len -- n )
  2swap 0 >r
  begin 2over search
  while 2over nip /string
        r> 1+ >r
  repeat 2drop 2drop r> ;

s" the three truths" s" th" str-count .    \ 3
s" ababababab" s" abab" str-count .     \ 2

Fortran

Works with: Fortran version 90 and later
program Example
  implicit none
  integer :: n
  
  n = countsubstring("the three truths", "th")
  write(*,*) n
  n = countsubstring("ababababab", "abab")
  write(*,*) n
  n = countsubstring("abaabba*bbaba*bbab", "a*b")
  write(*,*) n
 
contains

function countsubstring(s1, s2) result(c)
  character(*), intent(in) :: s1, s2
  integer :: c, p, posn
 
  c = 0
  if(len(s2) == 0) return
  p = 1
  do 
    posn = index(s1(p:), s2)
    if(posn == 0) return
    c = c + 1
    p = p + posn + len(s2) - 1
  end do
end function
end program
Output:
3
2
2

FreeBASIC

' FB 1.05.0 Win64

Function countSubstring(s As String, search As String) As Integer
  If s = "" OrElse search = "" Then Return 0
  Dim As Integer count = 0, length = Len(search)
  For i As Integer = 1 To Len(s)
    If Mid(s, i, length) = Search Then
      count += 1
      i += length - 1
    End If
  Next
  Return count
End Function

Print countSubstring("the three truths","th")
Print countSubstring("ababababab","abab")
Print countSubString("zzzzzzzzzzzzzzz", "z")
Print
Print "Press any key to quit"
Sleep
Output:
 3
 2
 15

FunL

import util.Regex

def countSubstring( str, substr ) = Regex( substr ).findAllMatchIn( str ).length()

println( countSubstring("the three truths", "th") )
println( countSubstring("ababababab", "abab") )
Output:
3
2

Fōrmulæ

Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text. Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for storage and transfer purposes more than visualization and edition.

Programs in Fōrmulæ are created/edited online in its website.

In this page you can see and run the program(s) related to this task and their results. You can also change either the programs or the parameters they are called with, for experimentation, but remember that these programs were created with the main purpose of showing a clear solution of the task, and they generally lack any kind of validation.

FutureBasic

window 1

local fn CountSubstring( string as CFStringRef, substring as CFStringRef ) as long
end fn = fn ArrayCount( fn StringComponentsSeparatedByString( string, substring ) ) - 1

print fn CountSubstring( @"the three truths", @"th" )
print fn CountSubstring( @"ababababab", @"abab" )

HandleEvents
Output:
3
2

Gambas

Translation of: FreeBASIC
Public Sub Main() 
  
  Print countSubstring("the three truths", "th") 
  Print countSubstring("ababababab", "abab") 
  Print countSubString("zzzzzzzzzzzzzzz", "z")

End 

Function countSubstring(s As String, search As String) As Integer 

  If s = "" Or search = "" Then Return 0 
  Dim count As Integer = 0, length As Integer = Len(search) 
  For i As Integer = 1 To Len(s) 
    If Mid(s, i, length) = Search Then 
      count += 1 
      i += length - 1 
    End If 
  Next 
  Return count 

End Function
Output:
Same as FreeBASIC entry.

Go

Using strings.Count() method:

package main
import (
        "fmt"
        "strings"
)

func main() {
        fmt.Println(strings.Count("the three truths", "th")) // says: 3
        fmt.Println(strings.Count("ababababab", "abab"))     // says: 2
}

Groovy

Solution, uses the Groovy "find" operator (=~), and the Groovy-extended Matcher property "count":

println (('the three truths' =~ /th/).count)
println (('ababababab' =~ /abab/).count)
println (('abaabba*bbaba*bbab' =~ /a*b/).count)
println (('abaabba*bbaba*bbab' =~ /a\*b/).count)
Output:
3
2
9
2

Haskell

Text-based solution

import Data.Text hiding (length)

-- Return the number of non-overlapping occurrences of sub in str.
countSubStrs str sub = length $ breakOnAll (pack sub) (pack str)

main = do
  print $ countSubStrs "the three truths" "th"
  print $ countSubStrs "ababababab" "abab"
Output:
3
2

Alternatively, in a language built around currying, it might make more sense to reverse the suggested order of arguments.

{-# LANGUAGE OverloadedStrings #-}

import Data.Text hiding (length)

--------- COUNT OF SUBSTRING INSTANCES IN A STRING -------

countAll :: Text -> Text -> Int
countAll needle haystack =
  length
    (breakOnAll needle haystack)

--------------------------- TEST -------------------------
main :: IO ()
main =
  print $
    countAll "ab"
      <$> [ "ababababab",
            "abelian absurdity",
            "babel kebab"
          ]
Output:
[5,2,2]

List-based solution

Even though list-based strings are not "the right" way of representing texts, the problem of counting subsequences in a list is generally useful.

count :: Eq a => [a] -> [a] -> Int
count []  = error "empty substring"
count sub = go
  where
    go = scan sub . dropWhile (/= head sub)
    scan _ [] = 0
    scan [] xs = 1 + go xs
    scan (x:xs) (y:ys) | x == y    = scan xs ys
                       | otherwise = go ys
Output:
λ> count "th" "the three truths"
3
λ> count "abab" "ababababab"
2
λ> count [2,3] [1,2,1,2,3,4,3,2,3,4,3,2]
2
λ> count "123456" $ foldMap show [1..1000000]
7

List-based solution using Data.List

The following solution is almost two times faster than the previous one.

import Data.List (tails, stripPrefix)
import Data.Maybe (catMaybes)

count :: Eq a => [a] -> [a] -> Int
count sub = length . catMaybes . map (stripPrefix sub) . tails

Icon and Unicon

procedure main()
every A := ![ ["the three truths","th"], ["ababababab","abab"] ] do        
   write("The string ",image(A[2])," occurs as a non-overlapping substring ",
         countSubstring!A , " times in ",image(A[1]))
end   

procedure countSubstring(s1,s2) #: return count of non-overlapping substrings
c := 0
s1 ? while tab(find(s2)) do {
   move(*s2)
   c +:= 1
   }
return c
end
Output:
The string "th" occurs as a non-overlapping substring 3 times in "the three truths"
The string "abab" occurs as a non-overlapping substring 2 times in "ababababab"

J

require'strings'
countss=: #@] %~ #@[ - [ #@rplc '';~]

In other words: find length of original string, replace the string to be counted with the empty string, find the difference in lengths and divide by the length of the string to be counted.

Example use:

   'the three truths' countss 'th'
3
   'ababababab' countss 'abab'
2

Java

Using regular expression pattern matching

int countSubstring(String string, String substring) {
    substring = Pattern.quote(substring);
    Pattern pattern = Pattern.compile(substring);
    Matcher matcher = pattern.matcher(string);
    int count = 0;
    while (matcher.find())
        count++;
    return count;
}


Works with: Java version 1.5+

The "remove and count the difference" method:

public class CountSubstring {
	public static int countSubstring(String subStr, String str){
		return (str.length() - str.replace(subStr, "").length()) / subStr.length();
	}
	
	public static void main(String[] args){
		System.out.println(countSubstring("th", "the three truths"));
		System.out.println(countSubstring("abab", "ababababab"));
		System.out.println(countSubstring("a*b", "abaabba*bbaba*bbab"));
	}
}
Output:
3
2
2
Works with: Java version 1.5+

The "split and count" method:

import java.util.regex.Pattern;

public class CountSubstring {
	public static int countSubstring(String subStr, String str){
		// the result of split() will contain one more element than the delimiter
		// the "-1" second argument makes it not discard trailing empty strings
		return str.split(Pattern.quote(subStr), -1).length - 1;
	}
	
	public static void main(String[] args){
		System.out.println(countSubstring("th", "the three truths"));
		System.out.println(countSubstring("abab", "ababababab"));
		System.out.println(countSubstring("a*b", "abaabba*bbaba*bbab"));
	}
}
Output:
3
2
2

Manual looping

public class CountSubstring {
	public static int countSubstring(String subStr, String str){
		int count = 0;
		for (int loc = str.indexOf(subStr); loc != -1;
		     loc = str.indexOf(subStr, loc + subStr.length()))
			count++;
		return count;
	}
	
	public static void main(String[] args){
		System.out.println(countSubstring("th", "the three truths"));
		System.out.println(countSubstring("abab", "ababababab"));
		System.out.println(countSubstring("a*b", "abaabba*bbaba*bbab"));
	}
}
Output:
3
2
2

JavaScript

Using regexes:

function countSubstring(str, subStr) {
    var matches = str.match(new RegExp(subStr, "g"));
    return matches ? matches.length : 0;
}

Using 'split' and ES6 notation:

const countSubString = (str, subStr) => str.split(subStr).length - 1;

jq

Using regexes (available in jq versions after June 19, 2014):

def countSubstring(sub):
  [match(sub; "g")] | length;
Example:
"the three truths" | countSubstring("th")

Julia

Built-in Function

matchall(r::Regex, s::String[, overlap::Bool=false]) -> Vector{String}

   Return a vector of the matching substrings from eachmatch.

Main

ts = ["the three truths", "ababababab"]
tsub = ["th", "abab"]

println("Test of non-overlapping substring counts.")
for i in 1:length(ts)
    print(ts[i], " (", tsub[i], ") => ")
    println(length(matchall(Regex(tsub[i]), ts[i])))
end
println()
println("Test of overlapping substring counts.")
for i in 1:length(ts)
    print(ts[i], " (", tsub[i], ") => ")
    println(length(matchall(Regex(tsub[i]), ts[i], true)))
end
Output:
Test of non-overlapping substring counts.
the three truths (th) => 3
ababababab (abab) => 2

Test of overlapping substring counts.
the three truths (th) => 3
ababababab (abab) => 4

K

The dyadic verb _ss gives the positions of substring y in string x.

  "the three truths" _ss "th"
0 4 13

  #"the three truths" _ss "th"
3

  "ababababab" _ss "abab"
0 4

  #"ababababab" _ss "abab"
2

Klingphix

include ..\Utilitys.tlhy

:count %s !s
    0 >ps
    [ps> 1 + >ps
     $s len nip + snip nip] [$s find dup] while
    drop drop ps>
;

"the three truths" "th" count ?
"ababababab" "abab" count ?

" " input

Other solution

include ..\Utilitys.tlhy

:count "- " convert "-" 2 tolist split len nip ;

"the three truths" "th" count ?
"ababababab" "abab" count ?

" " input
Output:
3
2

Kotlin

// version 1.0.6

fun countSubstring(s: String, sub: String): Int = s.split(sub).size - 1

fun main(args: Array<String>) {
    println(countSubstring("the three truths","th"))
    println(countSubstring("ababababab","abab"))
    println(countSubstring("",""))
}
Output:
3
2
1

Lambdatalk

{def countSubstring 
 {def countSubstring.r
  {lambda {:n :i :acc :s}
   {if {>= :i :n}
    then :acc
    else {countSubstring.r :n
                           {+ :i 1} 
                           {if {W.equal? {W.get :i :s} ⫖}
                            then {+ :acc 1}
                            else :acc}
                           :s} }}}
 {lambda {:w :s}
  {countSubstring.r {W.length :s} 0 0
                    {S.replace \s by  in 
                    {S.replace :w by  in :s}}}}}
-> countSubstring

{countSubstring th the three truths}
-> 3
{countSubstring ab ababa}
-> 2
{countSubstring aba ababa} 
-> 1

langur

writeln len indices "th", "the three truths"
writeln len indices "abab", "ababababab"
Output:
3
2

Lasso

define countSubstring(str::string, substr::string)::integer => {
	local(i = 1, foundpos = -1, found = 0)
	while(#i < #str->size && #foundpos != 0) => {
		protect => {
			handle_error => { #foundpos = 0 }
			#foundpos = #str->find(#substr, -offset=#i)
		}
		if(#foundpos > 0) => {
			#found += 1
			#i = #foundpos + #substr->size
		else
			#i++
		}
	}
	return #found
}
define countSubstring_bothways(str::string, substr::string)::integer => {
	local(found = countSubstring(#str,#substr))
	#str->reverse
	local(found2 = countSubstring(#str,#substr))
	#found > #found2 ? return #found | return #found2
}
countSubstring_bothways('the three truths','th')
//3
countSubstring_bothways('ababababab','abab')
//2

Liberty BASIC

print countSubstring( "the three truths", "th")
print countSubstring( "ababababab", "abab")
end

function countSubstring( a$, s$)
    c =0
    la =len( a$)
    ls =len( s$)
    for i =1 to la -ls
        if mid$( a$, i, ls) =s$ then c =c +1: i =i +ls -1
    next i
    countSubstring =c
end function

Logtalk

Using atoms for string representation:

:- object(counting).

    :- public(count/3).

    count(String, SubString, Count) :-
        count(String, SubString, 0, Count).

    count(String, SubString, Count0, Count) :-
        (   sub_atom(String, Before, Length, After, SubString) ->
            Count1 is Count0 + 1,
            Start is Before + Length,
            sub_atom(String, Start, After, 0, Rest),
            count(Rest, SubString, Count1, Count)
        ;   Count is Count0
        ).

:- end_object.
Output:
| ?- counting::count('the three truths', th, N).
N = 3
yes

| ?- counting::count(ababababab, abab, N).
N = 2
yes

Lua

Solution 1:

function countSubstring(s1, s2)
    return select(2, s1:gsub(s2, ""))
end

print(countSubstring("the three truths", "th"))
print(countSubstring("ababababab", "abab"))
3
2


Solution 2:

function countSubstring(s1, s2)
    local count = 0
    for eachMatch in s1:gmatch(s2) do 
        count = count + 1 
    end
    return count
end

print(countSubstring("the three truths", "th"))
print(countSubstring("ababababab", "abab"))
3
2

MACRO-11

        .TITLE  CSUBS
        .MCALL  .TTYOUT,.EXIT        
CSUBS:: JMP     DEMO
        
        ; COUNT SUBSTRINGS R1 IN R0 
COUNT:  CLR     R2
        BR      4$
1$:     MOV     R0,R3   
        MOV     R1,R4   
2$:     CMPB    (R3)+,(R4)+
        BEQ     2$
        TSTB    -(R4)
        BNE     3$
        INC     R2
        DEC     R3
        MOV     R3,R0
        BR      4$
3$:     INC     R0
4$:     TSTB    (R0)
        BNE     1$
        RTS     PC
        
        ; TEST EXAMPLES
DEMO:   MOV     #ST1,R0
        MOV     #SU1,R1
        JSR     PC,1$
        MOV     #ST2,R0
        MOV     #SU2,R1
        JSR     PC,1$
        .EXIT 
1$:     JSR     PC,COUNT
        ADD     #60,R2
        MOVB    R2,3$
        MOV     #3$,R2
2$:     MOVB    (R2)+,R0
        .TTYOUT
        BNE     2$
        RTS     PC
3$:     .BYTE   0,15,12,0

ST1:    .ASCIZ  /THE THREE TRUTHS/
SU1:    .ASCIZ  /TH/
ST2:    .ASCIZ  /ABABABABAB/
SU2:    .ASCIZ  /ABAB/
        .END    CSUBS
Output:
3
2

Maple

f:=proc(s::string,c::string,count::nonnegint) local n;
     n:=StringTools:-Search(c,s);
     if n>0 then 1+procname(s[n+length(c)..],c,count);
     else 0; end if;
end proc:

f("the three truths","th",0);

f("ababababab","abab",0);
Output:
                                      3

                                      2

Mathematica / Wolfram Language

StringPosition["the three truths","th",Overlaps->False]//Length
3
StringPosition["ababababab","abab",Overlaps->False]//Length
2

MATLAB / Octave

  % Count occurrences of a substring without overlap
  length(findstr("ababababab","abab",0))
  length(findstr("the three truths","th",0))

  % Count occurrences of a substring with overlap
  length(findstr("ababababab","abab",1))
Output:
>>   % Count occurrences of a substring without overlap
>>   length(findstr("ababababab","abab",0))
ans =  2
>>   length(findstr("the three truths","th",0))
ans =  3
>>   % Count occurrences of a substring with overlap
>>   length(findstr("ababababab","abab",1))
ans =  4
>> 

Maxima

scount(e, s) := block(
   [n: 0, k: 1],
   while integerp(k: ssearch(e, s, k)) do (n: n + 1, k: k + 1),
   n
)$

scount("na", "banana");
2

MiniScript

string.count = function(s)
    return self.split(s).len - 1
end function

print "the three truths".count("th")
print "ababababab".count("abab")
Output:
3
2

Mirah

import java.util.regex.Pattern
import java.util.regex.Matcher

#The "remove and count the difference" method
def count_substring(pattern:string, source:string)
    (source.length() - source.replace(pattern, "").length()) / pattern.length()
end
 
puts count_substring("th", "the three truths")      # ==> 3
puts count_substring("abab", "ababababab")          # ==> 2
puts count_substring("a*b", "abaabba*bbaba*bbab")   # ==> 2


# The "split and count" method
def count_substring2(pattern:string, source:string)
    # the result of split() will contain one more element than the delimiter
	# the "-1" second argument makes it not discard trailing empty strings
    source.split(Pattern.quote(pattern), -1).length - 1
end

puts count_substring2("th", "the three truths")      # ==> 3
puts count_substring2("abab", "ababababab")          # ==> 2
puts count_substring2("a*b", "abaabba*bbaba*bbab")   # ==> 2


# This method does a match and counts how many times it matches
def count_substring3(pattern:string, source:string)
    result = 0
    Matcher m = Pattern.compile(Pattern.quote(pattern)).matcher(source);
    while (m.find())
        result = result + 1
    end
    result
end

puts count_substring3("th", "the three truths")      # ==> 3
puts count_substring3("abab", "ababababab")          # ==> 2
puts count_substring3("a*b", "abaabba*bbaba*bbab")   # ==> 2

Miranda

main :: [sys_message]
main = [Stdout (show (countSubstring "the three truths" "th") ++ "\n"),
        Stdout (show (countSubstring "ababababab" "abab") ++ "\n")]

countSubstring :: [*]->[*]->num
countSubstring str ss
    = 0, if str = []
    = 1 + countSubstring (drop len str) ss, if match
    = countSubstring (tl str) ss, otherwise
      where len   = #ss
            match = take len str = ss
Output:
3
2

Nanoquery

Translation of: Java
def countSubstring(str, subStr)
        return int((len(str) - len(str.replace(subStr, ""))) / len(subStr))
end

Nemerle

Translation of: F#
using System.Console;

module CountSubStrings
{
    CountSubStrings(this text : string, target : string) : int
    {
        match (target) {
            |"" => 0
            |_ => (text.Length - text.Replace(target, "").Length) / target.Length
        }
    }
    
    Main() : void
    {
        def text1 = "the three truths";
        def target1 = "th";
        def text2 = "ababababab";
        def target2 = "abab";
        
        WriteLine($"$target1 occurs $(text1.CountSubStrings(target1)) times in $text1");
        WriteLine($"$target2 occurs $(text2.CountSubStrings(target2)) times in $text2");
    }
}
Output:
th occurs 3 times in the three truths
abab occurs 2 times in ababababab

NetRexx

NetRexx provides the string.countstr(needle) built-in function:

/* NetRexx */
options replace format comments java crossref symbols nobinary

-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
method countSubstring(inStr, findStr) public static
  return inStr.countstr(findStr)

-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
method main(args = String[]) public static
  strings = ''
  find = 'FIND'
  ix = 0
  ix = ix + 1; strings[0] = ix; find[0] = ix; strings[ix] = 'the three truths'; strings[ix, find] = 'th'
  ix = ix + 1; strings[0] = ix; find[0] = ix; strings[ix] = 'ababababab';       strings[ix, find] = 'abab'
  
  loop ix = 1 to strings[0]
    str = strings[ix]
    fnd = strings[ix, find]
    say 'there are' countSubstring(str, fnd) 'occurences of "'fnd'" in "'str'"'
    end ix
  
  return
Output:
there are 3 occurences of "th" in "the three truths"
there are 2 occurences of "abab" in "ababababab"

NewLISP

; file:   stringcount.lsp
; url:    http://rosettacode.org/wiki/Count_occurrences_of_a_substring
; author: oofoe 2012-01-29

; Obvious (and non-destructive...)

; Note that NewLISP performs an /implicit/ slice on a string or list
; with this form "(start# end# stringorlist)". If the end# is omitted,
; the slice will go to the end of the string. This is handy here to
; keep removing the front part of the string as it gets matched.

(define (scount needle haystack)
  (let ((h (copy haystack)) ; Copy of haystack string.
        (i 0)               ; Cursor.
        (c 0))              ; Count of occurences.

    (while (setq i (find needle h))
      (inc c)
      (setq h ((+ i (length needle)) h)))

    c))                     ; Return count.

; Tricky -- Uses functionality from replace function to find all
; non-overlapping occurrences, replace them, and return the count of
; items replaced in system variable $0.

(define (rcount needle haystack)
  (replace needle haystack "X") $0)

; Test

(define (test f needle haystack)
  (println "Found " (f needle haystack) 
           " occurences of '" needle "' in '" haystack "'."))

(dolist (f (list scount rcount))
        (test f "glart" "hinkerpop")
        (test f "abab"  "ababababab")
        (test f "th"    "the three truths")
        (println)
        )

(exit)
Output:
Found 0 occurences of 'glart' in 'hinkerpop'.
Found 2 occurences of 'abab' in 'ababababab'.
Found 3 occurences of 'th' in 'the three truths'.

Found 0 occurences of 'glart' in 'hinkerpop'.
Found 2 occurences of 'abab' in 'ababababab'.
Found 3 occurences of 'th' in 'the three truths'.

Nim

import strutils

proc count(s, sub: string): int =
  var i = 0
  while true:
    i = s.find(sub, i)
    if i < 0:
      break
    i += sub.len # i += 1 for overlapping substrings
    inc result

echo count("the three truths","th")

echo count("ababababab","abab")
Output:
3
2

Objective-C

The "split and count" method:

@interface NSString (CountSubstrings)
- (NSUInteger)occurrencesOfSubstring:(NSString *)subStr;
@end

@implementation NSString (CountSubstrings)
- (NSUInteger)occurrencesOfSubstring:(NSString *)subStr {
  return [[self componentsSeparatedByString:subStr] count] - 1;
}
@end

int main(int argc, const char *argv[]) {
  @autoreleasepool {

    NSLog(@"%lu", [@"the three truths" occurrencesOfSubstring:@"th"]);
    NSLog(@"%lu", [@"ababababab" occurrencesOfSubstring:@"abab"]);
    NSLog(@"%lu", [@"abaabba*bbaba*bbab" occurrencesOfSubstring:@"a*b"]);
 
  }
  return 0;
}
Output:
3
2
2


The "remove and count the difference" method:

@interface NSString (CountSubstrings)
- (NSUInteger)occurrencesOfSubstring:(NSString *)subStr;
@end

@implementation NSString (CountSubstrings)
- (NSUInteger)occurrencesOfSubstring:(NSString *)subStr {
  return ([self length] - [[self stringByReplacingOccurrencesOfString:subStr withString:@""] length]) / [subStr length];
}
@end

int main(int argc, const char *argv[]) {
  @autoreleasepool {

    NSLog(@"%lu", [@"the three truths" occurrencesOfSubstring:@"th"]);
    NSLog(@"%lu", [@"ababababab" occurrencesOfSubstring:@"abab"]);
    NSLog(@"%lu", [@"abaabba*bbaba*bbab" occurrencesOfSubstring:@"a*b"]);

  }
  return 0;
}
Output:
3
2
2


Manual looping:

@interface NSString (CountSubstrings)
- (NSUInteger)occurrencesOfSubstring:(NSString *)subStr;
@end

@implementation NSString (CountSubstrings)
- (NSUInteger)occurrencesOfSubstring:(NSString *)subStr {
  NSUInteger count = 0;
  for (NSRange range = [self rangeOfString:subStr]; range.location != NSNotFound;
       range.location += range.length,
       range = [self rangeOfString:subStr options:0
                             range:NSMakeRange(range.location, [self length] - range.location)])
    count++;
  return count;
}
@end

int main(int argc, const char *argv[]) {
  @autoreleasepool {

    NSLog(@"%lu", [@"the three truths" occurrencesOfSubstring:@"th"]);
    NSLog(@"%lu", [@"ababababab" occurrencesOfSubstring:@"abab"]);
    NSLog(@"%lu", [@"abaabba*bbaba*bbab" occurrencesOfSubstring:@"a*b"]);

  }
  return 0;
}
Output:
3
2
2

OCaml

let count_substring str sub =
  let sub_len = String.length sub in
  let len_diff = (String.length str) - sub_len
  and reg = Str.regexp_string sub in
  let rec aux i n =
    if i > len_diff then n else
      try
        let pos = Str.search_forward reg str i in
        aux (pos + sub_len) (succ n)
      with Not_found -> n
  in
  aux 0 0

let () =
  Printf.printf "count 1: %d\n" (count_substring "the three truth" "th");
  Printf.printf "count 2: %d\n" (count_substring "ababababab" "abab");
;;

Oforth

: countSubString(s, sub)
   0 1 while(sub swap s indexOfAllFrom dup notNull) [ sub size +  1 under+ ] 
   drop ;
Output:
countSubString("the three truths", "th") println
3
countSubString("ababababab", "abab") println
2

ooRexx

 bag="the three truths"
 x="th"
 say left(bag,30) left(x,15) 'found' bag~countstr(x)

 bag="ababababab"
 x="abab"
 say left(bag,30) left(x,15) 'found' bag~countstr(x)

 -- can be done caselessly too
 bag="abABAbaBab"
 x="abab"
 say left(bag,30) left(x,15) 'found' bag~caselesscountstr(x)
Output:
the three truths               th              found 3
ababababab                     abab            found 2
abABAbaBab                     abab            found 2

PARI/GP

subvec(v,u)={
	my(i=1,s);
	while(i+#u<=#v,
		for(j=1,#u,
			if(v[i+j-1]!=u[j], i++; next(2))
		);
		s++;
		i+=#u
	);
	s
};
substr(s1,s2)=subvec(Vec(s1),Vec(s2));
substr("the three truths","th")
substr("ababababab","abab")
Output:
%1 = 3
%2 = 2

Pascal

See Delphi

Perl

sub countSubstring {
  my $str = shift;
  my $sub = quotemeta(shift);
  my $count = () = $str =~ /$sub/g;
  return $count;
#  or return scalar( () = $str =~ /$sub/g );
}
  
print countSubstring("the three truths","th"), "\n"; # prints "3"
print countSubstring("ababababab","abab"), "\n"; # prints "2"

Phix

sequence tests = {{"the three truths","th"},
                  {"ababababab","abab"},
                  {"ababababab","aba"},
                  {"ababababab","ab"},
                  {"ababababab","a"},
                  {"ababababab",""}}
integer start, count
string test, substring
for i=1 to length(tests) do
    start = 1
    count = 0
    {test, substring} = tests[i]
    while 1 do
        start = match(substring,test,start)
        if start=0 then exit end if
        start += length(substring)
        count += 1
    end while
    printf(1,"The string \"%s\" occurs as a non-overlapping substring %d times in \"%s\"\n",{substring,count,test})
end for
Output:
The string "th" occurs as a non-overlapping substring 3 times in "the three truths"
The string "abab" occurs as a non-overlapping substring 2 times in "ababababab"
The string "aba" occurs as a non-overlapping substring 2 times in "ababababab"
The string "ab" occurs as a non-overlapping substring 5 times in "ababababab"
The string "a" occurs as a non-overlapping substring 5 times in "ababababab"
The string "" occurs as a non-overlapping substring 0 times in "ababababab"

PHP

<?php
echo substr_count("the three truths", "th"), PHP_EOL; // prints "3"
echo substr_count("ababababab", "abab"), PHP_EOL; // prints "2"

Picat

Picat has a predicate for searching for substrings (find/4) but on backtracking (e.g. in a findall/2 wrapper) it yields all overlapping substrings which is not correct in this task.

So we have to roll our own. Here are two versions.

Recursion

count_substrings_rec(S, SB) = C => 
    count_rec(S,SB,0,C).

count_rec([],_SB,Count,Count).
count_rec(SBRest,SB,Count0,Count) :-
  SBRest = SB ++ Rest, % "split" into substring and the rest of the string
  count_rec(Rest,SB,Count0+1,Count).
count_rec([T|Rest],SB,Count0,Count) :-
  T != SB, % this character is not a substring
  count_rec(Rest,SB,Count0,Count).

Iterative

Iterative version using find/4 (wrap with once/1 to avoid backtracking).

Translation of: Euphoria
count_substrings_find(S, SB) = C =>
   SLen = S.len,
   Count = 0,
   From = 1,
   while (From <= SLen)
     (
     once(find(slice(S,From),SB,_From2,To)) -> 
       Count := Count + 1,
       From := From + To 
     ;
       From := From + 1
     )
   end,
   C = Count.

The time differences between these two versions are quite large which is shown in a benchmark of searching the substring "ab" in a string of 100 000 random characters from the set of "abcde":

  • count_substrings_rec/2: 0.009s
  • count_substrings_find: 0.501s

PicoLisp

(de countSubstring (Str Sub)
   (let (Cnt 0  H (chop Sub))
      (for (S (chop Str)  S  (cdr S))
         (when (head H S)
            (inc 'Cnt)
            (setq S (map prog2 H S)) ) )
      Cnt ) )

Test:

: (countSubstring "the three truths" "th")
-> 3

: (countSubstring "ababababab" "abab")
-> 2

Pike

write("%d %d\n",
      String.count("the three truths", "th"),
      String.count("ababababab", "abab"));
Output:
3 2

PL/I

cnt: procedure options (main);
   declare (i, tally) fixed binary;
   declare (text, key) character (100) varying;

   get edit (text) (L); put skip data (text);
   get edit (key)  (L); put skip data (key);

   tally = 0; i = 1;
   do until (i = 0);
      i = index(text, key, i);
      if i > 0 then do; tally = tally + 1; i = i + length(key); end;
   end;
   put skip list (tally);
end cnt;

Output for the two specified strings is as expected.

Output:
for the following data
TEXT='AAAAAAAAAAAAAAA';
KEY='AA';
        7 

PL/M

100H:
/* CP/M CALLS */
BDOS: PROCEDURE (FN, ARG); DECLARE FN BYTE, ARG ADDRESS; GO TO 5; END BDOS;
EXIT: PROCEDURE; CALL BDOS(0, 0); END EXIT;
PRINT: PROCEDURE (S); DECLARE S ADDRESS; CALL BDOS(9, S); END PRINT;

/* PRINT A NUMBER */
PRINT$NUMBER: PROCEDURE (N);
    DECLARE S (8) BYTE INITIAL ('.....',13,10,'$');
    DECLARE (P, N) ADDRESS, C BASED P BYTE;
    P = .S(5);
DIGIT:
    P = P - 1;
    C = N MOD 10 + '0';
    N = N / 10;
    IF N > 0 THEN GO TO DIGIT;
    CALL PRINT(P);
END PRINT$NUMBER;

/* COUNT OCCURRENCES OF SUBSTRING IN STRING */
COUNT$SUBSTRING: PROCEDURE (STR, MATCH) ADDRESS;
    DECLARE (STR, MATCH) ADDRESS, C BASED STR BYTE;
    DECLARE (SP, MP) ADDRESS, (SC BASED SP, MC BASED MP) BYTE;
    DECLARE COUNT ADDRESS;
    COUNT = 0;
    DO WHILE C <> '$';
        SP = STR;
        MP = MATCH;
        DO WHILE SC = MC;
            SP = SP + 1;
            MP = MP + 1;
        END;
        IF MC = '$' THEN DO;
            STR = SP;
            COUNT = COUNT + 1;
        END;
        ELSE DO;
            STR = STR + 1;
        END;
    END;
    RETURN COUNT;
END COUNT$SUBSTRING;

CALL PRINT$NUMBER(COUNT$SUBSTRING(.'THE THREE TRUTHS$', .'TH$')); /* PRINTS 3 */
CALL PRINT$NUMBER(COUNT$SUBSTRING(.'ABABABABAB$', .'ABAB$')); /* PRINTS 2 */
CALL PRINT$NUMBER(COUNT$SUBSTRING(.'CAT$', .'DOG$')); /* PRINTS 0 */

CALL EXIT;
EOF
Output:
3
2
0

PowerBASIC

Works with: PB/Win
Works with: PB/CC

Windows versions of PowerBASIC (at least since PB/Win 7, and possibly earlier) provide the TALLY function, which does exactly what this task requires (count non-overlapping substrings).

PB/DOS can use the example under BASIC, above.

Note that while this example is marked as working with PB/Win, the PRINT statement would need to be replaced with MSGBOX, or output to a file. (PB/Win does not support console output.)

FUNCTION PBMAIN () AS LONG
    PRINT "the three truths, th:", TALLY("the three truths", "th")
    PRINT "ababababab, abab:", TALLY("ababababab", "abab")
END FUNCTION
Output:
the three truths, th:        3
ababababab, abab:            2

PowerShell

Works with: PowerShell version 4.0
[regex]::Matches("the three truths", "th").count

Output:

3
[regex]::Matches("ababababab","abab").count

Output:

2

Prolog

Works with: SWI-Prolog version 7

Using SWI-Prolog's string facilities (this solution is very similar to the Logtalk solution that uses sub_atom/5):

count_substring(String, Sub, Total) :-
    count_substring(String, Sub, 0, Total).

count_substring(String, Sub, Count, Total) :-
    ( substring_rest(String, Sub, Rest)
    ->
        succ(Count, NextCount),
        count_substring(Rest, Sub, NextCount, Total)
    ;
        Total = Count
    ).

substring_rest(String, Sub, Rest) :-
    sub_string(String, Before, Length, Remain, Sub),
    DropN is Before + Length,
    sub_string(String, DropN, Remain, 0, Rest).

Usage:

?- count_substring("the three truths","th",X).
X = 3.

?- count_substring("ababababab","abab",X).
X = 2.

version using DCG

Works with: SWI-Prolog version 7.6.4
:- system:set_prolog_flag(double_quotes,chars) .

occurrences(TARGETz0,SUBSTRINGz0,COUNT)
:-
prolog:phrase(occurrences(SUBSTRINGz0,0,COUNT),TARGETz0)
.

occurrences("",_,_)
-->
! ,
{ false }
.

occurrences(SUBSTRINGz0,COUNT0,COUNT)
-->
SUBSTRINGz0 ,
! ,
{ COUNT1 is COUNT0 + 1 } ,
occurrences(SUBSTRINGz0,COUNT1,COUNT)
.

occurrences(SUBSTRINGz0,COUNT0,COUNT)
-->
[_] ,
! ,
occurrences(SUBSTRINGz0,COUNT0,COUNT)
.

occurrences(_SUBSTRINGz0_,COUNT,COUNT)
-->
!
.
Output:
/*
?- occurrences("the three truths","th",COUNT) .
COUNT = 3.

?- occurrences("the three truths","",COUNT) .
false .

?- occurrences("ababababab","abab",COUNT) .
COUNT = 2.

?- occurrences("ababababab","",COUNT) .
false .

?- 
*/

PureBasic

a = CountString("the three truths","th")
b = CountString("ababababab","abab")        
; a = 3
; b = 2

Python

>>> "the three truths".count("th")
3
>>> "ababababab".count("abab")
2

Quackery

Quackery does not come equipped with a "find substring m within string n" function, but one is defined in The Book of Quackery, as a demonstration of creating a finite state machine in Quackery. It is reproduced here with permission.

  [ [] 95 times
    [ i^ space +
      join ] ]         constant is alphabet     (         --> $       )

  [ [ 2dup != while
      -1 split drop
      swap 1 split
      unrot drop again ]
    drop size ]                 is overlap      (     [ [ --> n       )

  [ temp put [] swap
    alphabet witheach
      [ over -1 poke
        over overlap
        dup temp share
        = if negate
        swap dip join ]
    drop temp release ]         is eachend      (     [ n --> [       )

  [ [] swap
    dup temp put
    size times
      [ temp share
        i 1+ split drop
        temp share size eachend
        nested swap join ]
    temp release ]              is buildfsm     (       $ --> [       )

  [ dup [] = iff -1
    else
      [ behead
        dup carriage = if
          [ drop space ]
        space - ]
    swap ]                      is nextcharn    (       $ --> n $     )

  [ swap dup size
    swap temp put
    swap 0
    [ over swap peek
      temp take
      nextcharn
      temp put
      dup 0 < iff
        [ 2drop 0 ] done
      peek
      dup 0 < until ]
    nip
    temp take size - + ]        is usefsm       (     $ [ --> n       )

  [ over size 0 = iff
      [ 2drop 0 ]
    else
      [ swap buildfsm 
        usefsm ] ]              is find$        (     $ $ --> n       )

find$ builds a finite state machine to search for m, (an O(m³) operation), then uses it to search in n with O(n). Rather than use find$, and repeatedly build the same fsm, we will define a word findall$ which returns a nest (i.e. list) of positions of m within n. (It actually returns the positions of the end of the substring, relative to (for the first instance) the start of the string, or (for subsequent instances) the end of the previous instance of the substring.)

  [ over size 0 = iff
      [ 2drop [] ] done
    [] unrot
    swap buildfsm
    [ 2dup usefsm
      rot 2dup found while
      dip [ over size + ]
      dip 
        [ rot over join 
          unrot ]
      swap split 
      nip swap again ]
     2drop drop ]               is findall$    (     $ $ --> [       )
Output:

Testing findall$ in the Quackery shell.

O> $ "th" $ "the three truths" tuck findall$
... witheach [ split swap echo$ cr ] 
... echo$
... 
th
e th
ree truth
s
Stack empty.

Finally we can use findall$ to fulfil the task requirements.

  [ swap findall$ size ]        is occurences  (     $ $ --> n       )

  $ "the three truths" $ "th" occurences echo cr
  $ "ababababab" $ "abab" occurences echo cr
Output:
3
2

R

The fixed parameter (and, in stringr, the function of the same name) is used to specify a search for a fixed string. Otherwise, the search pattern is interpreted as a POSIX regular expression. PCRE is also an option: use the perl parameter or function.

count = function(haystack, needle)
   {v = attr(gregexpr(needle, haystack, fixed = T)[[1]], "match.length")
    if (identical(v, -1L)) 0 else length(v)}

print(count("hello", "l"))
Library: stringr
library(stringr)
print(str_count("hello", fixed("l")))

Racket

(define count-substring 
  (compose length regexp-match*))
> (count-substring "th" "the three truths")
3
> (count-substring "abab" "ababababab")
2

Raku

(formerly Perl 6)

sub count-substring($big, $little) { +$big.comb: / :r $little / }
 
say count-substring("the three truths", "th"); # 3
say count-substring("ababababab", "abab");     # 2
 
say count-substring(123123123, 12);            # 3

The :r adverb makes the regex "ratchet forward" and skip any overlapping matches. .comb - when given a Regex as an argument - returns instances of that substring. Also, prefix + forces numeric context in Raku (it's a no-op in Perl 5). For the built in listy types that is the same as calling .elems method. One other style point: we now tend to prefer hyphenated names over camelCase.

Red

Red []

count-occurrences: function [string substring] [
    length? parse string [collect [some [keep substring to substring]]]
]

test-case-1: "the three truths"
test-case-2: "ababababab"

print [test-case-1 "-" count-occurrences test-case-1 "th"]
print [test-case-2 "-" count-occurrences test-case-2 "abab"]
Output:
the three truths - 3
ababababab - 2

Refal

$ENTRY Go {
    = <Prout <Count ('th') 'the three truths'>>
      <Prout <Count ('abab') 'abababab'>>;
};

Count {
    (e.item) e.item e.rest = <+ 1 <Count (e.item) e.rest>>;
    (e.item) s.x e.rest = <Count (e.item) e.rest>;
    (e.item) = 0;
};
Output:
3
2

REXX

Some older REXXes don't have the built-in function   countstr,   so one is included within the REXX program as a function.

The   countstr   subroutine (below) mimics the BIF in newer REXXes   (except for error checking).

Either of the first two strings may be null.

The third argument is optional and is the   start position   to start counting   (the default is   1,   meaning the first character).
If specified, it must be a positive integer   (and it may exceed the length of the 1st string).

The third argument was added here to be compatible with the newer REXXes BIF.

No checks are made (in the   countstr   subroutine) for:

  •   missing arguments
  •   too many arguments
  •   if   start   is a positive integer (when specified)
/*REXX program counts the  occurrences  of a (non─overlapping)  substring  in a string. */
w=.                                                                 /*max. width so far.*/
bag= 'the three truths'    ;      x= "th"       ;        call showResult
bag= 'ababababab'          ;      x= "abab"     ;        call showResult
bag= 'aaaabacad'           ;      x= "aa"       ;        call showResult
bag= 'abaabba*bbaba*bbab'  ;      x= "a*b"      ;        call showResult
bag= 'abaabba*bbaba*bbab'  ;      x= " "        ;        call showResult
bag=                       ;      x= "a"        ;        call showResult
bag=                       ;      x=            ;        call showResult
bag= 'catapultcatalog'     ;      x= "cat"      ;        call showResult
bag= 'aaaaaaaaaaaaaa'      ;      x= "aa"       ;        call showResult
exit                                             /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
countstr:   procedure;  parse arg haystack,needle,start;      if start==''  then start=1
            width=length(needle)
                                  do $=0  until p==0;         p=pos(needle,haystack,start)
                                  start=width + p                    /*prevent overlaps.*/
                                  end   /*$*/
            return $                                                 /*return the count.*/
/*──────────────────────────────────────────────────────────────────────────────────────*/
showResult: if w==. then do;                   w=30        /*W:  largest haystack width.*/
                         say center('haystack',w)  center('needle',w%2)  center('count',5)
                         say left('', w, "═")      left('', w%2, "═")    left('', 5, "═")
                         end

            if bag==''  then bag= " (null)"                /*handle displaying of nulls.*/
            if   x==''  then   x= " (null)"                /*   "        "      "   "   */
            say left(bag, w)           left(x, w%2)            center(countstr(bag, x), 5)
            return

output   when using the default (internal) inputs:

           haystack                needle      count
══════════════════════════════ ═══════════════ ═════
the three truths               th                3
ababababab                     abab              2
aaaabacad                      aa                2
abaabba*bbaba*bbab             a*b               2
abaabba*bbaba*bbab                               0
 (null)                        a                 0
 (null)                         (null)           1
catapultcatalog                cat               2
aaaaaaaaaaaaaa                 aa                7

Ring

aString = "Ring Welcome Ring to the Ring Ring Programming Ring Language Ring"
bString = "Ring"
see count(aString,bString)

func count cString,dString
     sum = 0
     while substr(cString,dString) > 0
           sum++
           cString = substr(cString,substr(cString,dString)+len(string(sum)))
     end
     return sum

Output:

6

RPL

Works with: Halcyon Calc version 4.2.7
≪ DUP SIZE 1 - → str substr subsize 
   ≪ 0
      1 str SIZE subsize + FOR j
          str j DUP subsize + SUB
          IF substr == THEN
             1 +
             j subsize + 'j' STO
          END
     NEXT
≫ ≫ 'CNTSUB' STO
"the three truths" CNTSUB
"ababababab" CNTSUB
Output:
2: 3
1: 2

Ruby

def countSubstrings str, subStr
  str.scan(subStr).length
end

p countSubstrings "the three truths", "th"      #=> 3
p countSubstrings "ababababab", "abab"          #=> 2

String#scan returns an array of substrings, and Array#length (or Array#size) counts them.

Run BASIC

print countSubstring("the three truths","th")
print countSubstring("ababababab","abab")

FUNCTION countSubstring(s$,find$)
WHILE instr(s$,find$,i) <> 0 
  countSubstring = countSubstring + 1
  i = instr(s$,find$,i) + len(find$)
WEND
END FUNCTION
Output:
3
2

Rust

fn main() {
    println!("{}","the three truths".matches("th").count());
    println!("{}","ababababab".matches("abab").count());
}
Output:
3
2

Scala

Using Recursion

import scala.annotation.tailrec
def countSubstring(str1:String, str2:String):Int={
   @tailrec def count(pos:Int, c:Int):Int={
      val idx=str1 indexOf(str2, pos)
      if(idx == -1) c else count(idx+str2.size, c+1)
   }
   count(0,0)
}

Using Sliding

def countSubstring(str: String, sub: String): Int =
  str.sliding(sub.length).count(_ == sub)

Using Regular Expressions

def countSubstring( str:String, substr:String ) = substr.r.findAllMatchIn(str).length


println(countSubstring("ababababab", "abab"))
println(countSubstring("the three truths", "th"))
Output:
2
3

Scheme

Works with: Gauche Scheme
gosh> (use gauche.lazy)
#<undef>
gosh> (length (lrxmatch "th" "the three truths"))
3
gosh> (length (lrxmatch "abab" "ababababab"))
2

Seed7

$ include "seed7_05.s7i";

const func integer: countSubstring (in string: stri, in string: searched) is func
  result
    var integer: count is 0;
  local
    var integer: offset is 0;
  begin
    offset := pos(stri, searched);
    while offset <> 0 do
      incr(count);
      offset := pos(stri, searched, offset + length(searched));
    end while;
  end func;
 
const proc: main is func
  begin
    writeln(countSubstring("the three truths", "th"));
    writeln(countSubstring("ababababab", "abab"));
  end func;
Output:
3
2

SETL

program count_overlapping_substrings;
    tests := [["the three truths", "th"], ["ababababab", "abab"]];
    loop for [s, subs] in tests do
        print("'" + subs + "' in '" + s + "': "
              + str countSubs(s, subs));
    end loop;

    proc countSubs(s, subs);
        count := 0;
        loop while s(subs) /= om do
            s(subs) := "";
            count +:= 1;
        end loop;
        return count;
    end proc;
end program;
Output:
'th' in 'the three truths': 3
'abab' in 'ababababab': 2

SenseTalk

Simply stated:

put the number of occurrences of "th" in "the three truths" --> 3
put the number of occurrences of "abab" in "ababababab" -- > 2

User-created function:

put countSubstring("aaaaa","a") // 5
put countSubstring("abababa","aba") // 2

function countSubstring mainString, subString
	return number of occurrences of subString in mainString
end countSubstring

Sidef

Built-in:

say "the three truths".count("th");
say "ababababab".count("abab");

User-created function:

func countSubstring(s, ss) {
    var re = Regex.new(ss.escape, 'g');      # 'g' for global
    var counter = 0;
    while (s =~ re) { ++counter };
    return counter;
}

say countSubstring("the three truths","th");
say countSubstring("ababababab","abab");
Output:
3
2

Simula

BEGIN
  
    INTEGER PROCEDURE COUNTSUBSTRING(T,TSUB); TEXT T,TSUB;
    BEGIN
        INTEGER N,I;
        I := 1;
        WHILE I+TSUB.LENGTH-1 <= T.LENGTH DO
          IF T.SUB(I,TSUB.LENGTH) = TSUB THEN
          BEGIN
              N := N+1;
              I := I+MAX(TSUB.LENGTH,1);
          END ELSE I := I+1;
        COUNTSUBSTRING:= N;
    END COUNTSUBSTRING;

    OUTINT(COUNTSUBSTRING("THE THREE TRUTHS", "TH"),0);
    OUTIMAGE;
    OUTINT(COUNTSUBSTRING("ABABABABAB", "ABAB"),0);
    OUTIMAGE;
END.
Output:
3
2

Smalltalk

Works with: Smalltalk/X
Transcript showCR:('the three truths' occurrencesOfString:'th').
Transcript showCR:('ababababab' occurrencesOfString:'abab')
Output:
3
2

SNOBOL4

        DEFINE("countSubstring(t,s)")

        OUTPUT = countSubstring("the three truths","th")
        OUTPUT = countSubstring("ababababab","abab")

        :(END)
countSubstring t ARB s = :F(RETURN)
        countSubstring = countSubstring + 1 :(countSubstring)
END
3
2

Standard ML

fun count_substrings (str, sub) =
  let
    fun aux (str', count) =
      let
        val suff = #2 (Substring.position sub str')
      in
        if Substring.isEmpty suff then
       	  count
        else
          aux (Substring.triml (size sub) suff, count + 1)
      end
  in
    aux (Substring.full str, 0)
  end;

print (Int.toString (count_substrings ("the three truths", "th")) ^ "\n");
print (Int.toString (count_substrings ("ababababab", "abab")) ^ "\n");
print (Int.toString (count_substrings ("abaabba*bbaba*bbab", "a*b")) ^ "\n");

Stata

function strcount(s, x) {
	n = 0
	k = 1-(i=strlen(x))
	do {
		if (k = ustrpos(s, x, k+i)) n++
	} while(k)
	return(n)
}

strcount("peter piper picked a peck of pickled peppers", "pe")
  5

strcount("ababababab","abab")
  2

Swift

import Foundation

func countSubstring(str: String, substring: String) -> Int {
  return str.components(separatedBy: substring).count - 1
}

print(countSubstring(str: "the three truths", substring: "th"))
print(countSubstring(str: "ababababab", substring: "abab"))
Output:
3
2

Tcl

The regular expression engine is ideal for this task, especially as the ***= prefix makes it interpret the rest of the argument as a literal string to match:

proc countSubstrings {haystack needle} {
    regexp -all ***=$needle $haystack
}
puts [countSubstrings "the three truths" "th"]
puts [countSubstrings "ababababab" "abab"]
puts [countSubstrings "abaabba*bbaba*bbab" "a*b"]
Output:
3
2
2

Transd

#lang transd

MainModule: {
    countSubstring: (λ s String() sub String() 
        (with n 0 pl 0
            (while (> (= pl (find s sub pl)) -1) 
                (+= pl (size sub)) (+= n 1))
            (lout n))
    ),
    _start: (λ 
        (countSubstring "the three truths" "th")
        (countSubstring "ababababab" "abab")
    )
}
Output:
3
2

TUSCRIPT

$$ MODE TUSCRIPT, {}
occurences=COUNT ("the three truths",  ":th:")
occurences=COUNT ("ababababab",   ":abab:")
occurences=COUNT ("abaabba*bbaba*bbab",":a\*b:")
Output:
3
2
2

TXR

@(next :args)
@(do (defun count-occurrences (haystack needle)
       (for* ((occurrences 0)
              (old-pos 0)
              (new-pos (search-str haystack needle old-pos nil)))
             (new-pos occurrences)
             ((inc occurrences)
              (set old-pos (+ new-pos (length needle)))
              (set new-pos (search-str haystack needle old-pos nil))))))
@ndl
@hay
@(output)
@(count-occurrences hay ndl) occurrences(s) of @ndl inside @hay
@(end)
$ ./txr count-occurrences.txr "baba" "babababa"
2 occurence(s) of baba inside babababa
$ ./txr count-occurrences.txr "cat" "catapultcatalog"
2 occurence(s) of cat inside catapultcatalog

UNIX Shell

Works with: Bash
#!/bin/bash

function countString(){
	input=$1
	cnt=0

	until [ "${input/$2/}" == "$input" ]; do
		input=${input/$2/}
		let cnt+=1
	done
	echo $cnt
}

countString "the three truths" "th"
countString "ababababab" "abab"
Output:
3
2

VBA

Function CountStringInString(stLookIn As String, stLookFor As String)
    CountStringInString = UBound(Split(stLookIn, stLookFor))
End Function

VBScript

user created function

Function CountSubstring(str,substr)
	CountSubstring = 0
	For i = 1 To Len(str)
		If Len(str) >= Len(substr) Then
			If InStr(i,str,substr) Then
				CountSubstring = CountSubstring + 1
				i = InStr(i,str,substr) + Len(substr) - 1
			End If
		Else
			Exit For
		End If
	Next
End Function

WScript.StdOut.Write CountSubstring("the three truths","th") & vbCrLf
WScript.StdOut.Write CountSubstring("ababababab","abab") & vbCrLf

Using built-in Regexp

Run it with CScript.

function CountSubstring(str,substr)
  with new regexp
     .pattern=substr
     .global=true
     set m=.execute(str)
  end with
  CountSubstring =m.count
end function  
WScript.StdOut.Writeline CountSubstring("the three truths","th") 
WScript.StdOut.Writeline CountSubstring("ababababab","abab")
Output:
3
2

Visual Basic .NET

Module Count_Occurrences_of_a_Substring
    Sub Main()
        Console.WriteLine(CountSubstring("the three truths", "th"))
        Console.WriteLine(CountSubstring("ababababab", "abab"))
        Console.WriteLine(CountSubstring("abaabba*bbaba*bbab", "a*b"))
        Console.WriteLine(CountSubstring("abc", ""))
    End Sub

    Function CountSubstring(str As String, substr As String) As Integer
        Dim count As Integer = 0
        If (Len(str) > 0) And (Len(substr) > 0) Then
            Dim p As Integer = InStr(str, substr)
            Do While p <> 0
                p = InStr(p + Len(substr), str, substr)
                count += 1
            Loop
        End If
        Return count
    End Function
End Module
Output:
3
2
2
0

V (Vlang)

fn main(){
    println('the three truths'.count('th'))
    println('ababababab'.count('abab'))
}
Output:
3
2

Wortel

@let {
  c &[s t] #!s.match &(t)g

  [[
    !!c "the three truths" "th"
    !!c "ababababab" "abab"
  ]]
}
Returns:
[3 2]

Wren

Library: Wren-pattern
Library: Wren-fmt
import "./pattern" for Pattern
import "./fmt" for Fmt

var countSubstring = Fn.new { |str, sub|
    var p = Pattern.new(sub)
    return p.findAll(str).count
}

var tests = [
    ["the three truths", "th"],
    ["ababababab", "abab"],
    ["abaabba*bbaba*bbab", "a*b"],
    ["aaaaaaaaaaaaaa", "aa"],
    ["aaaaaaaaaaaaaa", "b"],
]

for (test in tests) {
    var count = countSubstring.call(test[0], test[1])
    Fmt.print("$6s occurs $d times in $q.", Fmt.q(test[1]), count, test[0])
}
Output:
  "th" occurs 3 times in "the three truths".
"abab" occurs 2 times in "ababababab".
 "a*b" occurs 2 times in "abaabba*bbaba*bbab".
  "aa" occurs 7 times in "aaaaaaaaaaaaaa".
   "b" occurs 0 times in "aaaaaaaaaaaaaa".

Alternatively, using a library method (output same as before):

Library: Wren-str
import "./str" for Str
import "./fmt" for Fmt

var tests = [
    ["the three truths", "th"],
    ["ababababab", "abab"],
    ["abaabba*bbaba*bbab", "a*b"],
    ["aaaaaaaaaaaaaa", "aa"],
    ["aaaaaaaaaaaaaa", "b"],
]

for (test in tests) {
    var count = Str.occurs(test[0], test[1])
    Fmt.print("$6s occurs $d times in $q.", Fmt.q(test[1]), count, test[0])
}

XPL0

include c:\cxpl\codes;  \intrinsic 'code' declarations
string 0;               \use zero-terminated strings, instead of MSb terminated


func    StrNCmp(A, B, N); \Compare string A to string B up to N bytes long
\This returns:
\       >0 if A > B
\       =0 if A = B
\       <0 if A < B
char    A, B;           \strings to be compared
int     N;              \number of bytes to compare
int     I;
[for I:= 0 to N-1 do
    if A(I) # B(I) then
        return A(I) - B(I);
return 0;               \they're equal
];      \StrNCmp


func    StrLen(Str);    \Return the number of characters in an ASCIIZ string
char    Str;
int     I;
for I:= 0 to -1>>1-1 do
        if Str(I) = 0 then return I;


func    SubStr(A, B);   \Count number of times string B occurs in A
char    A, B;
int     LA, LB, C, I;
[LA:= StrLen(A);  LB:= StrLen(B);
C:= 0;  I:= 0;
while I < LA do
        if StrNCmp(B, A+I, LB) = 0 then [C:= C+1;  I:= I+LB]
        else I:= I+1;
return C;
];


[IntOut(0, SubStr("the three truths", "th"));  CrLf(0);
 IntOut(0, SubStr("ababababab", "abab"));  CrLf(0);
]
Output:
3
2
>>   % Count occurrences of a substring without overlap
>>   length(findstr("ababababab","abab",0))
ans =  2
>>   length(findstr("the three truths","th",0))
ans =  3
>>   % Count occurrences of a substring with overlap
>>   length(findstr("ababababab","abab",1))
ans =  4
>>

zkl

Two solutions:

fcn countSubstring(s,p){ pn:=p.len(); cnt:=n:=0;
   while(Void!=(n:=s.find(p,n))){cnt+=1; n+=pn}
   cnt
}
Translation of: J
fcn countSubstring(s,p){ (pl:=p.len()) and (s.len()-(s-p).len())/pl }
Output:
zkl: println(countSubstring("the three truths","th"))
3
zkl: println(countSubstring("ababababab","abab"))
2
zkl: println(countSubstring("ababababab","v"))
0

ZX Spectrum Basic

10 LET t$="ABABABABAB": LET p$="ABAB": GO SUB 1000
20 LET t$="THE THREE TRUTHS": LET p$="TH": GO SUB 1000
30 STOP 
1000 PRINT t$: LET c=0
1010 LET lp=LEN p$
1020 FOR i=1 TO LEN t$-lp+1
1030 IF (t$(i TO i+lp-1)=p$) THEN LET c=c+1: LET i=i+lp-1
1040 NEXT i
1050 PRINT p$;"=";c''
1060 RETURN