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).


Metrics: length

Sub-string search: Count occurrences of a substring

Multi-string operations: LCP, LCS, concatenation

Manipulation: reverse, lower- and uppercase

360 Assembly[edit]

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

Ada[edit]

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

ALGOL 68[edit]

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[edit]

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



AppleScript[edit]

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}

AutoHotkey[edit]

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[edit]

#
# 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

BASIC[edit]

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[edit]

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

Batch File[edit]

@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[edit]

      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"

Bracmat[edit]

  ( 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[edit]

#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++[edit]

#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

C#[edit]

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;
}
}


Clojure[edit]

Use a sequence of regexp matches to count occurrences.

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

COBOL[edit]

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[edit]

 
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[edit]

(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

D[edit]

void main() {
import std.stdio, std.algorithm;
 
"the three truths".count("th").writeln;
"ababababab".count("abab").writeln;
}
Output:
3
2

Delphi[edit]

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.

Déjà Vu[edit]

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

EchoLisp[edit]

 
;; 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
 

Eiffel[edit]

 
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[edit]

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[edit]

 
%% 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[edit]

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

EGL[edit]

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

Factor[edit]

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

Forth[edit]

: 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[edit]

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)
end do
end function
end program
Output:
3
2
2

F#[edit]

"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

FunL[edit]

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

Go[edit]

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[edit]

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[edit]

 
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

Icon and Unicon[edit]

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[edit]

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

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[edit]

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[edit]

Using regexes:

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

jq[edit]

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

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

Julia[edit]

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[edit]

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
 


Lasso[edit]

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[edit]

 
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[edit]

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[edit]

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

Maple[edit]

 
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[edit]

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

MATLAB / Octave[edit]

  % 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[edit]

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

Mirah[edit]

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
 

Nemerle[edit]

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[edit]

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[edit]

; 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[edit]

import strutils
 
proc count(s, sub): 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[edit]

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[edit]

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[edit]

 
: 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[edit]

 
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[edit]

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[edit]

See Delphi

Perl[edit]

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"

Perl 6[edit]

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

Note that in Perl 6 the /$little/ matches the variable literally, so there's no need to quote regex metacharacters. Also, prefix + forces numeric context in Perl 6 (it's a no-op in Perl 5). One other style point: we now tend to prefer hyphenated names over camelCase.

Phix[edit]

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[edit]

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

PicoLisp[edit]

(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

PL/I[edit]

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 

PowerBASIC[edit]

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

PureBasic[edit]

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

PowerShell[edit]

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

Output:

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

Output:

2

Prolog[edit]

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.
 

Python[edit]

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

R[edit]

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[edit]

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

REXX[edit]

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[edit]

 
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

Ruby[edit]

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[edit]

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

Scala[edit]

Using Recursion[edit]

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 Regular Expressions[edit]

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[edit]

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

Seed7[edit]

$ 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

Sidef[edit]

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

SNOBOL4[edit]

 
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[edit]

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");

TUSCRIPT[edit]

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

Tcl[edit]

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

TXR[edit]

@(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[edit]

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[edit]

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

VBScript[edit]

 
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
 
Output:
3
2

Wortel[edit]

@let {
c &[s t] #!s.match &(t)g
 
[[
 !!c "the three truths" "th"
 !!c "ababababab" "abab"
]]
}
Returns:
[3 2]

XPL0[edit]

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[edit]

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[edit]

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