Count in octal

From Rosetta Code
Task
Count in octal
You are encouraged to solve this task according to the task description, using any language you may know.
Task

Produce a sequential count in octal,   starting at zero,   and using an increment of a one for each consecutive number.

Each number should appear on a single line,   and the program should count until terminated,   or until the maximum value of the numeric type in use is reached.


Related task



0815[edit]

}:l:>     Start loop, enqueue Z (initially 0).
  }:o:    Treat the queue as a stack and
    <:8:= accumulate the octal digits
    /=>&~ of the current number.
  ^:o:

  <:0:-   Get a sentinel negative 1.
  &>@     Enqueue it between the digits and the current number.
  {       Dequeue the first octal digit.

  }:p:
    ~%={+ Rotate each octal digit into place and print it.
  ^:p:

  <:a:~$  Output a newline.
  <:1:x{+ Dequeue the current number and increment it.
^:l:

360 Assembly[edit]

The program uses one ASSIST macro (XPRNT) to keep the code as short as possible.

*        Octal                     04/07/2016
OCTAL    CSECT
         USING  OCTAL,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            "
         LA     R6,0               i=0
LOOPI    LR     R2,R6              x=i
         LA     R9,10              j=10
         LA     R4,PG+23           @pg
LOOP     LR     R3,R2              save x
         SLL    R2,29              shift left  32-3
         SRL    R2,29              shift right 32-3
         CVD    R2,DW              convert octal(j) to pack decimal 
         OI     DW+7,X'0F'         prepare unpack
         UNPK   0(1,R4),DW         packed decimal to zoned printable
         LR     R2,R3              restore x
         SRL    R2,3               shift right 3
         BCTR   R4,0               @pg=@pg-1
         BCT    R9,LOOP            j=j-1
         CVD    R2,DW              binary to pack decimal 
         OI     DW+7,X'0F'         prepare unpack
         UNPK   0(1,R4),DW         packed decimal to zoned printable
         CVD    R6,DW              convert i to pack decimal 
         MVC    ZN12,EM12          load mask
         ED     ZN12,DW+2          packed decimal (PL6) to char (CL12)
         MVC    PG(12),ZN12        output i
         XPRNT  PG,80              print buffer
         C      R6,=F'2147483647'  if i>2**31-1 (integer max)
         BE     ELOOPI             then exit loop on i
         LA     R6,1(R6)           i=i+1
         B      LOOPI              loop on i
ELOOPI   L      R13,4(0,R13)       epilog 
         LM     R14,R12,12(R13)    "
         XR     R15,R15            "
         BR     R14                exit
         LTORG  
PG       DC     CL80' '            buffer
DW       DS     0D,PL8             15num
ZN12     DS     CL12
EM12     DC     X'40',9X'20',X'2120'  mask CL12 11num
         YREGS
         END    OCTAL
Output:
           0 00000000000
           1 00000000001
           2 00000000002
           3 00000000003
           4 00000000004
           5 00000000005
           6 00000000006
           7 00000000007
           8 00000000010
           9 00000000011
          10 00000000012
          10 00000000012
          11 00000000013
...
  2147483640 17777777770
  2147483641 17777777771
  2147483642 17777777772
  2147483643 17777777773
  2147483644 17777777774
  2147483645 17777777775
  2147483646 17777777776
  2147483647 17777777777

6502 Assembly[edit]

Works with: [Easy6502]

Easy6502 can only output using a limited video memory or a hexdump. However the output is correct up to 2 octal digits.

define SRC_LO  $00
define SRC_HI  $01

define DEST_LO $02
define DEST_HI $03

define temp $04		;temp storage used by foo

;some prep work since easy6502 doesn't allow you to define arbitrary bytes before runtime.

SET_TABLE:
TXA
STA $1000,X
INX
BNE SET_TABLE	
;stores the identity table at memory address $1000-$10FF

CLEAR_TABLE:
LDA #0
STA $1200,X
INX
BNE CLEAR_TABLE
;fills the range $1200-$12FF with zeroes.


LDA #$10
STA SRC_HI
LDA #$00
STA SRC_LO
;store memory address $1000 in zero page

LDA #$12
STA DEST_HI
LDA #$00
STA DEST_LO
;store memory address $1200 in zero page


loop:
LDA (SRC_LO),y  ;load accumulator from memory address $1000+y
JSR foo		;convert accumulator to octal
STA (DEST_LO),y ;store accumulator in memory address $1200+y

INY
CPY #$40 
BCC loop
BRK

foo:
sta temp           ;store input temporarily
asl                ;bit shift, this places the top bit of the right nibble in the bottom of the left nibble.
pha                ;back this value up
    lda temp
    and #$07       ;take the original input and remove everything except the bottom 3 bits.
    sta temp       ;store it for later. What used to be stored here is no longer needed.
pla                ;get the pushed value back.
and #$F0           ;clear the bottom 4 bits.
ora temp           ;put the bottom 3 bits of the original input back.
and #$7F           ;clear bit 7.
rts
Output:
1200: 00 01 02 03 04 05 06 07 10 11 12 13 14 15 16 17 
1210: 20 21 22 23 24 25 26 27 30 31 32 33 34 35 36 37 
1220: 40 41 42 43 44 45 46 47 50 51 52 53 54 55 56 57 
1230: 60 61 62 63 64 65 66 67 70 71 72 73 74 75 76 77 

8080 Assembly[edit]

This assumes the CP/M operating system. The count will terminate after the largest unsigned 16-bit value is reached.

	;-------------------------------------------------------
	; useful equates
	;-------------------------------------------------------
bdos	equ	5	; CP/M BDOS entry
conout	equ	2	; BDOS console output function
cr	equ	13	; ASCII carriage return
lf	equ	10	; ASCII line feed
	;------------------------------------------------------
	; main code begins here
	;------------------------------------------------------
	org	100h	; start of tpa under CP/M
	lxi	h,0	; save CP/M's stack
	dad	sp
	shld	oldstk
	lxi	sp,stack ; set our own stack
	lxi	h,1	; start counting at 1
count:	call	putoct
	call	crlf
	inx	h
	mov	a,h	; check for overflow (hl = 0)
	ora	l
	jnz	count
	;
	; all finished. clean up and exit.
	;
	lhld	oldstk	; get CP/M's stack back
	sphl		; restore it
	ret		; exit to command prompt 
	;------------------------------------------------------
	; Octal output routine
	; entry:  hl = number to output on console in octal
	; this is a recursive routine and uses 6 bytes of stack
	; space for each digit
	;------------------------------------------------------
putoct: push	b
	push	d
	push	h
	mvi	b,3	; hl = hl >> 3
div2:	call	shlr	
	dcr	b
	jnz	div2
	mov	a,l	; test if hl = 0
	ora	h
	cnz	putoct  ; recursive call
	pop	h	; get unshifted hl back
	push	h
	mov	a,l	; get low byte
	ani	7	; a = a mod 8
	adi	'0'	; make printable
	call	putchr
	pop	h
	pop	d
	pop	b
	ret        
	;-------------------------------------------------------
	; logical right shift of 16-bit value in HL by one bit
	;-------------------------------------------------------
shlr:	ora	a	; clear carry
	mov	a,h	; begin with most significant byte
	rar		; bit 0 goes into carry
	mov	h,a	; put shifted byte back
	mov	a,l	; get least significant byte
	rar		; bit 0 of MSB has shifted in
	mov	l,a
	ret
	;------------------------------------------------------
	; output CRLF to console
	;------------------------------------------------------
crlf:	mvi	a,cr
	call	putchr
	mvi	a,lf
	call	putchr
	ret
	;------------------------------------------------------
	; Console output routine
	; print character in A register to console
        ; preserves BC, DE, and HL
	;------------------------------------------------------
putchr:	push	h
	push	d
	push	b
	mov	e,a	; character to E for CP/M
	mvi	c,conout
	call	bdos
	pop	b
	pop	d
	pop	h
	ret
	;-------------------------------------------------------
	;  data area
	;-------------------------------------------------------
oldstk:	dw	1
stack	equ	$+128	; 64 level stack
	;
	end
Output:

Showing the last 10 lines of the output.

1777766
1777767
1777770
1777771
1777772
1777773
1777774
1777775
1777776
1777777

AArch64 Assembly[edit]

Works with: as version Raspberry Pi 3B version Buster 64 bits
/* ARM assembly AARCH64 Raspberry PI 3B */
/*  program countOctal64.s   */
 
/*******************************************/
/* Constantes file                         */
/*******************************************/
/* for this file see task include a file in language AArch64 assembly*/
.include "../includeConstantesARM64.inc"
 
/*********************************/
/* Initialized data              */
/*********************************/
.data
sMessResult:        .ascii "Count : "
sMessValeur:        .fill 11, 1, ' '            // size => 11
szCarriageReturn:   .asciz "\n"
 
 
/*********************************/
/* UnInitialized data            */
/*********************************/
.bss  
/*********************************/
/*  code section                 */
/*********************************/
.text
.global main 
main:                                             // entry of program 
    mov x20,0                                     // loop indice
1:                                                // begin loop
    mov x0,x20
    ldr x1,qAdrsMessValeur
    bl conversion8                                // call conversion octal
    ldr x0,qAdrsMessResult
    bl affichageMess                              // display message
    add x20,x20,1
    cmp x20,64
    ble 1b
 
 
100:                                              // standard end of the program 
    mov x0,0                                      // return code
    mov x8,EXIT                                   // request to exit program
    svc 0                                         // perform the system call
 
qAdrsMessValeur:          .quad sMessValeur
qAdrszCarriageReturn:     .quad szCarriageReturn
qAdrsMessResult:          .quad sMessResult
 

/******************************************************************/
/*     Converting a register to octal                             */ 
/******************************************************************/
/* x0 contains value and x1 address area   */
/* x0 return size of result (no zero final in area) */
/* area size => 11 bytes          */
.equ LGZONECAL,   10
conversion8:
    stp x1,lr,[sp,-16]!            // save  registers
    mov x3,x1
    mov x2,LGZONECAL
 
1:                                 // start loop
    mov x1,x0
    lsr x0,x0,3                    // / by 8
    lsl x4,x0,3
    sub x1,x1,x4                   // compute remainder x1 - (x0 * 8)
    add x1,x1,48                   // digit
    strb w1,[x3,x2]                // store digit on area
    cmp x0,0                       // stop if quotient = 0 
    sub x4,x2,1
    csel x2,x4,x2,ne
    bne 1b                         // and loop
                                   // and move digit from left of area
    mov x4,0
2:
    ldrb w1,[x3,x2]
    strb w1,[x3,x4]
    add x2,x2,1
    add x4,x4,1
    cmp x2,LGZONECAL
    ble 2b
                                   // and move spaces in end on area
    mov x0,x4                      // result length 
    mov x1,' '                     // space
3:
    strb w1,[x3,x4]                // store space in area
    add x4,x4,1                       // next position
    cmp x4,LGZONECAL
    ble 3b                         // loop if x4 <= area size
 
100:

    ldp x1,lr,[sp],16              // restaur  2 registers
    ret                            // return to address lr x30
/********************************************************/
/*        File Include fonctions                        */
/********************************************************/
/* for this file see task include a file in language AArch64 assembly */
.include "../includeARM64.inc"

Action![edit]

PROC PrintOctal(CARD v)
  CHAR ARRAY a(6) 
  BYTE i=[0]

  DO 
    a(i)=(v&7)+'0
    i==+1
    v=v RSH 3
  UNTIL v=0
  OD

  DO
    i==-1
    Put(a(i))
  UNTIL i=0
  OD
RETURN

PROC Main()
  CARD i=[0]

  DO 
    PrintF("decimal=%U octal=",i)
    PrintOctal(i) PutE()
    i==+1
  UNTIL i=0
  OD
RETURN
Output:

Screenshot from Atari 8-bit computer

decimal=0 octal=0
decimal=1 octal=1
decimal=2 octal=2
decimal=3 octal=3
decimal=4 octal=4
...
decimal=3818 octal=7352
decimal=3819 octal=7353
decimal=3820 octal=7354
decimal=3821 octal=7355
decimal=3822 octal=7356
...

Ada[edit]

with Ada.Text_IO;

procedure Octal is
   package IIO is new Ada.Text_IO.Integer_IO(Integer);
begin
   for I in 0 .. Integer'Last loop
      IIO.Put(I, Base => 8);
      Ada.Text_IO.New_Line;
   end loop;
end Octal;

First few lines of Output:

       8#0#
       8#1#
       8#2#
       8#3#
       8#4#
       8#5#
       8#6#
       8#7#
      8#10#
      8#11#
      8#12#
      8#13#
      8#14#
      8#15#
      8#16#
      8#17#
      8#20#

Aime[edit]

integer o;

o = 0;    
do {     
    o_xinteger(8, o);
    o_byte('\n');
    o += 1;
} while (0 < o);

ALGOL 68[edit]

Works with: ALGOL 68G version Any - tested with release 1.18.0-9h.tiny.
#!/usr/local/bin/a68g --script #

INT oct width = (bits width-1) OVER 3 + 1;
main:
(
  FOR i TO 17 # max int # DO
    printf(($"8r"8r n(oct width)dl$, BIN i))
  OD
)

Output:

8r00000000001
8r00000000002
8r00000000003
8r00000000004
8r00000000005
8r00000000006
8r00000000007
8r00000000010
8r00000000011
8r00000000012
8r00000000013
8r00000000014
8r00000000015
8r00000000016
8r00000000017
8r00000000020
8r00000000021

ALGOL W[edit]

Algol W has built-in hexadecimal and decimal output, this implements octal output.

begin
    string(12) r;
    string(8)  octDigits;
    integer    number;
    octDigits := "01234567";
    number    := -1;
    while number < MAXINTEGER do begin
        integer    v, cPos;
        number := number + 1;
        v      := number;
        % build a string of octal digits in r, representing number %
        % Algol W uses 32 bit integers, so r should be big enough  %
        % the most significant digit is on the right               %
        cPos   := 0;
        while begin
            r( cPos // 1 ) := octDigits( v rem 8 // 1 );
            v :=  v div 8;
            ( v > 0 )
        end do begin            
            cPos := cPos + 1
        end while_v_gt_0;
        % show most significant digit on a newline %
        write( r( cPos // 1 ) );
        % continue the line with the remaining digits (if any) %
        for c := cPos - 1 step -1 until 0 do writeon( r( c // 1 ) )
    end while_r_lt_MAXINTEGER
end.
Output:
0
1
2
3
4
5
6
7
10
11
12
...

APL[edit]

Works with Dyalog APL. 100,000 is just an arbitrarily large number I chose.

10¨8¯1¨100000

ARM Assembly[edit]

Works with: as version Raspberry Pi
/* ARM assembly Raspberry PI  */
/*  program countoctal.s   */
 
/************************************/
/* Constantes                       */
/************************************/
.equ STDOUT, 1     @ Linux output console
.equ EXIT,   1     @ Linux syscall
.equ WRITE,  4     @ Linux syscall

/*********************************/
/* Initialized data              */
/*********************************/
.data
sMessResult:        .ascii "Count : "
sMessValeur:        .fill 11, 1, ' '            @ size => 11
szCarriageReturn:   .asciz "\n"


/*********************************/
/* UnInitialized data            */
/*********************************/
.bss  
/*********************************/
/*  code section                 */
/*********************************/
.text
.global main 
main:                                             @ entry of program 
    mov r4,#0                                     @ loop indice
1:                                                @ begin loop
    mov r0,r4
    ldr r1,iAdrsMessValeur
    bl conversion8                                @ call conversion octal
    ldr r0,iAdrsMessResult
    bl affichageMess                              @ display message
    add r4,#1
    cmp r4,#64
    ble 1b


100:                                              @ standard end of the program 
    mov r0, #0                                    @ return code
    mov r7, #EXIT                                 @ request to exit program
    svc #0                                        @ perform the system call
 
iAdrsMessValeur:          .int sMessValeur
iAdrszCarriageReturn:     .int szCarriageReturn
iAdrsMessResult:          .int sMessResult

/******************************************************************/
/*     display text with size calculation                         */ 
/******************************************************************/
/* r0 contains the address of the message */
affichageMess:
    push {r0,r1,r2,r7,lr}                          @ save  registres
    mov r2,#0                                      @ counter length 
1:                                                 @ loop length calculation 
    ldrb r1,[r0,r2]                                @ read octet start position + index 
    cmp r1,#0                                      @ if 0 its over 
    addne r2,r2,#1                                 @ else add 1 in the length 
    bne 1b                                         @ and loop 
                                                   @ so here r2 contains the length of the message 
    mov r1,r0                                      @ address message in r1 
    mov r0,#STDOUT                                 @ code to write to the standard output Linux 
    mov r7, #WRITE                                 @ code call system "write" 
    svc #0                                         @ call systeme 
    pop {r0,r1,r2,r7,lr}                           @ restaur des  2 registres */ 
    bx lr                                          @ return  
/******************************************************************/
/*     Converting a register to octal                             */ 
/******************************************************************/
/* r0 contains value and r1 address area   */
/* r0 return size of result (no zero final in area) */
/* area size => 11 bytes          */
.equ LGZONECAL,   10
conversion8:
    push {r1-r4,lr}                                 @ save registers 
    mov r3,r1
    mov r2,#LGZONECAL
 
1:                                                  @ start loop
    mov r1,r0
    lsr r0,#3                                       @ / by 8
    sub r1,r0,lsl #3                                @ compute remainder r1 - (r0 * 8)
    add r1,#48                                      @ digit
    strb r1,[r3,r2]                                 @ store digit on area
    cmp r0,#0                                       @ stop if quotient = 0 
    subne r2,#1                                     @ else previous position
    bne 1b                                          @ and loop
                                                    @ and move digit from left of area
    mov r4,#0
2:
    ldrb r1,[r3,r2]
    strb r1,[r3,r4]
    add r2,#1
    add r4,#1
    cmp r2,#LGZONECAL
    ble 2b
                                                      @ and move spaces in end on area
    mov r0,r4                                         @ result length 
    mov r1,#' '                                       @ space
3:
    strb r1,[r3,r4]                                   @ store space in area
    add r4,#1                                         @ next position
    cmp r4,#LGZONECAL
    ble 3b                                            @ loop if r4 <= area size
 
100:
    pop {r1-r4,lr}                                    @ restaur registres 
    bx lr                                             @return

Arturo[edit]

loop 1..40 'i ->
	print ["number in base 10:" pad to :string i 2 
		   "number in octal:" pad as.octal i 2]
Output:
number in base 10:  1 number in octal:  1 
number in base 10:  2 number in octal:  2 
number in base 10:  3 number in octal:  3 
number in base 10:  4 number in octal:  4 
number in base 10:  5 number in octal:  5 
number in base 10:  6 number in octal:  6 
number in base 10:  7 number in octal:  7 
number in base 10:  8 number in octal: 10 
number in base 10:  9 number in octal: 11 
number in base 10: 10 number in octal: 12 
number in base 10: 11 number in octal: 13 
number in base 10: 12 number in octal: 14 
number in base 10: 13 number in octal: 15 
number in base 10: 14 number in octal: 16 
number in base 10: 15 number in octal: 17 
number in base 10: 16 number in octal: 20 
number in base 10: 17 number in octal: 21 
number in base 10: 18 number in octal: 22 
number in base 10: 19 number in octal: 23 
number in base 10: 20 number in octal: 24 
number in base 10: 21 number in octal: 25 
number in base 10: 22 number in octal: 26 
number in base 10: 23 number in octal: 27 
number in base 10: 24 number in octal: 30 
number in base 10: 25 number in octal: 31 
number in base 10: 26 number in octal: 32 
number in base 10: 27 number in octal: 33 
number in base 10: 28 number in octal: 34 
number in base 10: 29 number in octal: 35 
number in base 10: 30 number in octal: 36 
number in base 10: 31 number in octal: 37 
number in base 10: 32 number in octal: 40 
number in base 10: 33 number in octal: 41 
number in base 10: 34 number in octal: 42 
number in base 10: 35 number in octal: 43 
number in base 10: 36 number in octal: 44 
number in base 10: 37 number in octal: 45 
number in base 10: 38 number in octal: 46 
number in base 10: 39 number in octal: 47 
number in base 10: 40 number in octal: 50

AutoHotkey[edit]

DllCall("AllocConsole")
Octal(int){
	While int
		out := Mod(int, 8) . out, int := int//8
	return out
}
Loop
{
	FileAppend, % Octal(A_Index) "`n", CONOUT$
	Sleep 200
}

AWK[edit]

The awk extraction and reporting language uses the underlying C library to provide support for the printf command. This enables us to use that function to output the counter value as octal:

BEGIN {
  for (l = 0; l <= 2147483647; l++) {
    printf("%o\n", l);
  }
}

BASIC[edit]

Some BASICs provide a built-in function to convert a number to octal, typically called OCT$.

Works with: QBasic
DIM n AS LONG
FOR n = 0 TO &h7FFFFFFF
    PRINT OCT$(n)
NEXT

However, many do not. For those BASICs, we need to write our own function.

Works with: Chipmunk Basic
WHILE ("" = INKEY$)
    PRINT Octal$(n)
    n = n + 1
WEND
END
FUNCTION Octal$(what)
    outp$ = ""
    w = what
    WHILE ABS(w) > 0
        o = w AND 7
        w = INT(w / 8)
        outp$ = STR$(o) + outp$
    WEND
    Octal$ = outp$
END FUNCTION

See also: BBC BASIC, Liberty BASIC, PureBasic, Run BASIC

Applesoft BASIC[edit]

10 N$ = "0"

100 O$ = N$
110 PRINT O$
120 N$ = ""
130 C = 1
140 FOR I = LEN(O$) TO 1 STEP -1
150     N = VAL(MID$(O$, I, 1)) + C
160     C = N >= 8
170     N$ = STR$(N - C * 8) + N$
180 NEXT I
190 IF C THEN N$ = "1" + N$
200 GOTO 100

BASIC256[edit]

valor = 0
do
	print ToOctal(valor)
	valor += 1
until valor = 0
end


Commodore BASIC[edit]

This example calculates the octal equivalent of the number and returns the octal equivalent in the form of a string.

Eventually, the number will reach 1,000,000,000 (one billion decimal) at which point the computer will express the value of n in exponential format, i.e. 1e+09 and will thus loose precision and stop counting.

Commodore BASIC has a little quirk where numeric values converted to a string also include a leading space for the possible negative sign; this is why the STR$ function is wrapped in a RIGHT$ function.

10 n=0
20 gosub 70
30 print oc$
40 n=n+1
50 get a$:if a$<>"q" then goto 20
60 end
70 oc$="":t=n
80 q=int(t/8)
90 r=t-(q*8)
100 oc$=left$(str$(n),1)+right$(str$(r),1)+oc$
110 if q<>0 then t=q:goto 80
120 return
Output:
0
1
2
3
4
5
6
7
10
11
12
13
14
15
17
20
21
22
23

User stopped count.

ready.
█


Sinclair ZX81 BASIC[edit]

The octal number is stored and manipulated as a string, meaning that even with only 1k of RAM the program shouldn't stop until the number gets to a couple of hundred digits long. I have not left it running long enough to find out exactly when it does run out of memory. The SCROLL statement is necessary: the ZX81 halts when the screen is full unless it is positively told to scroll instead.

 10 LET N$="0"
 20 SCROLL
 30 PRINT N$
 40 LET L=LEN N$
 50 LET N=VAL N$(L)+1
 60 IF N=8 THEN GOTO 90
 70 LET N$(L)=STR$ N
 80 GOTO 20
 90 LET N$(L)="0"
100 IF L=1 THEN GOTO 130
110 LET L=L-1
120 GOTO 50
130 LET N$="1"+N$
140 GOTO 20

uBasic/4tH[edit]

This routine allows for any base (up to 36) and also caters for negative numbers.

x = 1

Do
  Print Show(FUNC(_FNtobase(x, 8)))
  While Set (x, x+1)
Loop

End
      
_FNtobase  
  Param (2)                            ' convert A@ to string in base B@
  Local (2)                            ' digit C@ and string D@
                                       ' initialize, save sign
  d@ = Dup("") : Push a@ < 0 : a@ = Abs(a@)
      
  Do
    c@ = a@ % b@ : a@ = a@ / b@        ' extract digit and append
    d@ = Join (Char (Ord("0") + c@ + (7 * (c@ > 9))), d@)
  While a@ > 0                         ' something left to convert?
  Loop
      
  If Pop() Then d@ = Join ("-", d@)    ' apply sign if required
Return (d@)

Batch File[edit]

@echo off
:: {CTRL + C} to exit the batch file

:: Send incrementing decimal values to the :to_Oct function
set loop=0
:loop1
call:to_Oct %loop%
set /a loop+=1
goto loop1

:: Convert the decimal values parsed [%1] to octal and output them on a new line
:to_Oct
set todivide=%1
set "fulloct="

:loop2
set tomod=%todivide%
set /a appendmod=%tomod% %% 8
set fulloct=%appendmod%%fulloct%
if %todivide% lss 8 (
  echo %fulloct%
  exit /b
)
set /a todivide/=8
goto loop2
Output:
0
1
2
3
4
5
6
7
10
...

BBC BASIC[edit]

Terminate by pressing ESCape.

      N% = 0
      REPEAT
        PRINT FN_tobase(N%, 8, 0)
        N% += 1
      UNTIL FALSE
      END
      
      REM Convert N% to string in base B% with minimum M% digits:
      DEF FN_tobase(N%, B%, M%)
      LOCAL D%, A$
      REPEAT
        D% = N% MOD B%
        N% DIV= B%
        IF D%<0 D% += B% : N% -= 1
        A$ = CHR$(48 + D% - 7*(D%>9)) + A$
        M% -= 1
      UNTIL (N%=FALSE OR N%=TRUE) AND M%<=0
      =A$

bc[edit]

obase = 8			/* Output base is octal. */
for (num = 0; 1; num++) num	/* Loop forever, printing counter. */

The loop never stops at a maximum value, because bc uses arbitrary-precision integers.

BCPL[edit]

This will count up from 0 until the limit of the machine word.

get "libhdr"

let start() be
$(  let x = 0
    $(  writeo(x)
        wrch('*N')
        x := x + 1
    $) repeatuntil x = 0
$)

Befunge[edit]

This is almost identical to the Binary digits sample, except for the change of base and the source coming from a loop rather than a single input.

:0\55+\:8%68>*#<+#8\#68#%/#8:_$>:#,_$1+:0`!#@_

BQN[edit]

_while_ and Oct are snippets from BQNcrate. A more array-oriented approach is ⥊↕n⥊8, which produces all n-digit octal numbers instead of counting.

_while_←{𝔽⍟𝔾∘𝔽_𝕣_𝔾∘𝔽⍟𝔾𝕩}
Oct←8{⌽𝕗|⌊∘÷⟜𝕗⍟(↕1+·⌊𝕗⋆⁼1⌈⊢)}
{•Show Oct 𝕩, 𝕩+1} _while_ 1 0

Bracmat[edit]

Stops when the user presses Ctrl-C or when the stack overflows. The solution is not elegant, and so is octal counting.

  ( oct
  =   
    .     !arg:<8
        & (!arg:~<0|ERROR)
      | str$(oct$(div$(!arg.8)) mod$(!arg.8))
  )
& -1:?n
& whl'(1+!n:?n&out$(!n oct$!n));

Brainf***[edit]

+[            Start with n=1 to kick off the loop
[>>++++++++<< Set up {n 0 8} for divmod magic
[->+>-        Then
[>+>>]>       do
[+[-<+>]>+>>] the
<<<<<<]       magic
>>>+          Increment n % 8 so that 0s don't break things
>]            Move into n / 8 and divmod that unless it's 0
-<            Set up sentinel ‑1 then move into the first octal digit
[++++++++ ++++++++ ++++++++ Add 47 to get it to ASCII
 ++++++++ ++++++++ +++++++. and print it
[<]<]         Get to a 0; the cell to the left is the next octal digit
>>[<+>-]      Tape is {0 n}; make it {n 0}
>[>+]         Get to the ‑1
<[[-]<]       Zero the tape for the next iteration
++++++++++.   Print a newline
[-]<+]        Zero it then increment n and go again

C[edit]

#include <stdio.h>

int main()
{
        unsigned int i = 0;
        do { printf("%o\n", i++); } while(i);
        return 0;
}

C#[edit]

using System;

class Program
{
    static void Main()
    {
        var number = 0;
        do
        {
            Console.WriteLine(Convert.ToString(number, 8));
        } while (++number > 0);
    }
}

C++[edit]

This prevents an infinite loop by counting until the counter overflows and produces a 0 again. This could also be done with a for or while loop, but you'd have to print 0 (or the last number) outside the loop.

#include <iostream>

int main()
{
  unsigned i = 0;
  do
  {
    std::cout << std::oct << i << std::endl;
    ++i;
  } while(i != 0);
  return 0;
}

Clojure[edit]

(doseq [i (range)] (println (format "%o" i)))

COBOL[edit]

Translation of: Delphi
Works with: GNU Cobol version 2.0
       >>SOURCE FREE
IDENTIFICATION DIVISION.
PROGRAM-ID. count-in-octal.

ENVIRONMENT DIVISION.
CONFIGURATION SECTION.
REPOSITORY.
    FUNCTION dec-to-oct
    .
DATA DIVISION.
WORKING-STORAGE SECTION.
01  i                                   PIC 9(18).

PROCEDURE DIVISION.
    PERFORM VARYING i FROM 1 BY 1 UNTIL i = 0
        DISPLAY FUNCTION dec-to-oct(i)
    END-PERFORM
    .
END PROGRAM count-in-octal.


IDENTIFICATION DIVISION.
FUNCTION-ID. dec-to-oct.

DATA DIVISION.
LOCAL-STORAGE SECTION.
01  rem                                 PIC 9.

01  dec                                 PIC 9(18).

LINKAGE SECTION.
01  dec-arg                             PIC 9(18).

01  oct                                 PIC 9(18).

PROCEDURE DIVISION USING dec-arg RETURNING oct.
    MOVE dec-arg TO dec *> Copy is made to avoid modifying reference arg.
    PERFORM WITH TEST AFTER UNTIL dec = 0
        MOVE FUNCTION REM(dec, 8) TO rem
        STRING rem, oct DELIMITED BY SPACES INTO oct
        DIVIDE 8 INTO dec
    END-PERFORM
    .
END FUNCTION dec-to-oct.

CoffeeScript[edit]

n = 0

while true
  console.log n.toString(8)
  n += 1

Common Lisp[edit]

(loop for i from 0 do (format t "~o~%" i))

Component Pascal[edit]

BlackBox Component Builder

MODULE CountOctal;
IMPORT StdLog,Strings;

PROCEDURE Do*;
VAR
	i: INTEGER;
	resp: ARRAY 32 OF CHAR;
BEGIN
	FOR i := 0 TO 1000 DO
		Strings.IntToStringForm(i,8,12,' ',TRUE,resp);
		StdLog.String(resp);StdLog.Ln
	END
END Do;
END CountOctal.

Execute: ^Q CountOctal.Do
Output:

         0%8
         1%8
         2%8
         3%8
         4%8
         5%8
         6%8
         7%8
        10%8
        11%8
        12%8
        13%8
        14%8
        15%8
        16%8
        17%8
        20%8
        21%8
        22%8 

Cowgol[edit]

include "cowgol.coh";

typedef N is uint16; 

sub print_octal(n: N) is
    var buf: uint8[12];
    var p := &buf[11];
    [p] := 0;
    loop
        p := @prev p;
        [p] := '0' + (n as uint8 & 7);
        n := n >> 3;
        if n == 0 then break; end if;
    end loop;
    print(p);
end sub;

var n: N := 0;
loop
    print_octal(n);
    print_nl();
    n := n + 1;
    if n == 0 then break; end if; 
end loop;

Crystal[edit]

# version 0.21.1
# using unsigned 8 bit integer, range 0 to 255
 
(0_u8..255_u8).each { |i| puts i.to_s(8) }
Output:
0
1
2
3
4
5
6
7
10
11
12
...
374
375
376
377

D[edit]

void main() {
    import std.stdio;

    ubyte i;
    do writefln("%o", i++);
    while(i);
}

Dc[edit]

Named Macro[edit]

A simple infinite loop and octal output will do.

8o0[p1+lpx]dspx

Anonymous Macro[edit]

Needs r (swap TOS and NOS):

8 o 0 [ r p 1 + r dx ] dx

Pushing/poping TOS to a named stack can be used instead of swaping:

8 o 0 [ S@ p 1 + L@ dx ] dx

DCL[edit]

$ i = 0
$ loop:
$  write sys$output f$fao( "!OL", i )
$  i = i + 1
$  goto loop
Output:
00000000000
00000000001
00000000002
...
17777777777
20000000000
20000000001
...
37777777777
00000000000
00000000001
...

Delphi[edit]

program CountingInOctal;

{$APPTYPE CONSOLE}

uses SysUtils;

function DecToOct(aValue: Integer): string;
var
  lRemainder: Integer;
begin
  Result := '';
  repeat
    lRemainder := aValue mod 8;
    Result := IntToStr(lRemainder) + Result;
    aValue := aValue div 8;
  until aValue = 0;
end;

var
  i: Integer;
begin
  for i := 0 to 20 do
    WriteLn(DecToOct(i));
end.

EDSAC order code[edit]

Uses 17-bit integers, maximum 2^16 - 1 (177777 octal). It would take the original EDSAC 18 or 19 hours to exhaust these, so there is not much point in extending to 35-bit integers.

[Count in octal, for Rosetta Code.
 EDSAC program, Initial Orders 2.]

[Subroutine to print 17-bit non-negative integer in octal,
   with suppression of leading zeros.
 Input: 0F = number (not preserved)
 Workspace: 0D, 4F, 5F]
            T64K GK  [load at location 64]
            A3F T28@ [plant return link as usual]
            T4D      [clear whole of 4D including sandwich bit]
            A2F      [load 0...010 binary (permanently in 2F)]
            T4F      [(1) marker bit (2) flag to test for leading 0]
            AF       [load number]
            R2F      [shift 3 right]
            A4D      [add marker bit]
            TD       [store number and marker in 0D]
            H29@     [mask to isolate 3-bit octal digit]
[Loop to print digits]
     [10]   T5F      [clear acc]
            C1F      [top 5 bits of acc = octal digit]
            U5F      [to 5F for printing]
            S4F      [subtract flag to test for leading 0]
            G18@     [skip printing if so]
            O5F      [print digit]
            T5F      [clear acc]
            T4F      [flag = 0, so future 0's are not skipped]
     [18]   T5F      [clear acc]
            AD       [load number + marker bit, as shifted]
            L2F      [shift left 3 more]
            TD       [store back]
            AF       [has marker reached sign bit yet?]
            E10@     [loop back if not]
[Last digit separately, in case input = 0]
            T5F      [clear acc]
            C1F T5F O5F
     [28]   ZF       [(planted) jump back to caller]
     [29]   UF       [mask, 001110...0 binary]

[Main routine]
            T96K GK  [load at location 96]
[Constants]
      [0]   PD       [1]
      [1]   #F       [set figures mode]
      [2]   @F       [carriage return]
      [3]   &F       [line feed]
      [4]   K4096F   [null char]
[Variable]
      [5]   PF       [number to be printed]
[Enter with acc = 0]
      [6]   O1@      [set teleprinter to figures]
      [7]   U5@      [update number, initially 0]
            TF       [also to 0F for printing]
      [9]   A9@ G64F [call print soubroutine]
            O2@ O3@  [print CR, LF]
            A5@ A@   [load number, add 1]
            E7@      [loop until number overflows and becomes negative]
            O4@      [done; print null to flush teleprinter buffer]
            ZF       [halt the machine]
            E6Z      [define entry point]
            PF       [acc = 0 on entry]
[end]
Output:
0
1
2
3
4
5
6
7
10
[...]
177775
177776
177777

Elixir[edit]

Stream.iterate(0,&(&1+1)) |> Enum.each(&IO.puts Integer.to_string(&1,8))

or

Stream.unfold(0, fn n ->
  IO.puts Integer.to_string(n,8)
  {n,n+1}
end) |> Stream.run

or

f = fn ff,i -> :io.fwrite "~.8b~n", [i]; ff.(ff, i+1) end
f.(f, 0)

Emacs Lisp[edit]

Displays in the message area interactively, or to standard output under -batch.

(dotimes (i most-positive-fixnum) ;; starting from 0
  (message "%o" i))

Erlang[edit]

The fun is copied from Integer sequence#Erlang. I changed the display format.

F = fun(FF, I) -> io:fwrite("~.8B~n", [I]), FF(FF, I + 1) end.

Use like this:

F( F, 0 ).

Euphoria[edit]

integer i
i = 0
while 1 do
    printf(1,"%o\n",i)
    i += 1
end while

Output:

...
6326
6327
6330
6331
6332
6333
6334
6335
6336
6337

F#[edit]

let rec countInOctal num : unit =
  printfn "%o" num
  countInOctal (num + 1)

countInOctal 1

Factor[edit]

USING: kernel math prettyprint ;
0 [ dup .o 1 + t ] loop

Forth[edit]

Using INTS from Integer sequence#Forth

: octal ( -- )  8 base ! ;  \ where unavailable

octal ints

Fortran[edit]

Works with: Fortran version 95 and later
program Octal
  implicit none
  
  integer, parameter :: i64 = selected_int_kind(18)
  integer(i64) :: n = 0
  
! Will stop when n overflows from
! 9223372036854775807 to -92233720368547758078 (1000000000000000000000 octal)
  do while(n >= 0)
    write(*, "(o0)") n
    n = n + 1
  end do
end program

FreeBASIC[edit]

' FB 1.05.0 Win64

Dim ub As UByte = 0 ' only has a range of 0 to 255
Do
   Print Oct(ub, 3)
   ub += 1
Loop Until ub = 0 ' wraps around to 0 when reaches 256
Print
Print "Press any key to quit"
Sleep

Frink[edit]

i = 0
while true
{
    println[i -> octal]
    i = i + 1
}

Futhark[edit]

Futhark cannot print. Instead we produce an array of integers that look like octal numbers when printed in decimal.

fun octal(x: int): int =
  loop ((out,mult,x) = (0,1,x)) = while x > 0 do
    let digit = x % 8
    let out = out + digit * mult
    in (out, mult * 10, x / 8)
  in out

fun main(n: int): [n]int =
  map octal (iota n)

FutureBasic[edit]

window 1, @"Count in Octal"

defstr word
dim as short i
text ,,,,, 50

print @"dec",@"oct"
for i = 0 to 25
  print i,oct(i)
next

HandleEvents

Output:

dec	oct
0	00
1	01
2	02
3	03
4	04
5	05
6	06
7	07
8	10
9	11
10	12
11	13
12	14
13	15
14	16
15	17
16	20
17	21
18	22
19	23
20	24
21	25
22	26
23	27
24	30
25	31

Go[edit]

package main

import (
    "fmt"
    "math"
)

func main() {
    for i := int8(0); ; i++ {
        fmt.Printf("%o\n", i)
        if i == math.MaxInt8 {
            break
        }
    }
}

Output:

0
1
2
3
4
5
6
7
10
11
12
...
175
176
177

Note that to use a different integer type, code must be changed in two places. Go has no way to query a type for its maximum value. Example:

func main() {
    for i := uint16(0); ; i++ {  // type specified here
        fmt.Printf("%o\n", i)
        if i == math.MaxUint16 { // maximum value for type specified here
            break
        }
    }
}

Output:

...
177775
177776
177777

Note also that if floating point types are used for the counter, loss of precision will prevent the program from from ever reaching the maximum value. If you stretch interpretation of the task wording "maximum value" to mean "maximum value of contiguous integers" then the following will work:

import "fmt"

func main() {
    for i := 0.; ; {
        fmt.Printf("%o\n", int64(i))
        /* uncomment to produce example output
        if i == 3 {
            i = float64(1<<53 - 4) // skip to near the end
            fmt.Println("...")
        } */
        next := i + 1
        if next == i {
            break
        }
        i = next
    }
}

Output, with skip uncommented:

0
1
2
3
...
377777777777777775
377777777777777776
377777777777777777
400000000000000000

Big integers have no maximum value, but the Go runtime will panic when memory allocation fails. The deferred recover here allows the program to terminate silently should the program run until this happens.

import (
    "big"
    "fmt"
)

func main() {
    defer func() {
        recover()
    }()
    one := big.NewInt(1)
    for i := big.NewInt(0); ; i.Add(i, one) {
        fmt.Printf("%o\n", i)
    }
}

Output:

0
1
2
3
4
5
6
7
10
11
12
13
14
...

Groovy[edit]

Size-limited solution:

println 'decimal  octal'
for (def i = 0; i <= Integer.MAX_VALUE; i++) {
    printf ('%7d  %#5o\n', i, i)
}

Unbounded solution:

println 'decimal  octal'
for (def i = 0g; true; i += 1g) {
    printf ('%7d  %#5o\n', i, i)
}

Output:

decimal  octal
      0     00
      1     01
      2     02
      3     03
      4     04
      5     05
      6     06
      7     07
      8    010
      9    011
     10    012
     11    013
     12    014
     13    015
     14    016
     15    017
     16    020
     17    021
...

Haskell[edit]

import Numeric (showOct)

main :: IO ()
main =
  mapM_
    (putStrLn . flip showOct "")
    [1 .. maxBound :: Int]

Icon and Unicon[edit]

link convert   # To get exbase10 method

procedure main()
    limit := 8r37777777777
    every write(exbase10(seq(0)\limit, 8))
end

J[edit]

Solution:

   disp=.([echo) ' '(-.~":)8&#.inv
   (1+disp)^:_]0x

The full result is not displayable, by design. This could be considered a bug, but is an essential part of this task. Here's how it starts:

   (1+disp)^:_]0x
0
1
2
3
4
5
6
7
10
11
...

The important part of this code is 8&#.inv which converts numbers from internal representation to a sequence of base 8 digits. (We then convert this sequence to characters and remove the delimiting spaces - this gives us the octal values we want to display.)

So then we define disp as a word which displays its argument in octal and returns its argument as its result (unchanged).

Finally, the ^:_ clause tells J to repeat this function forever, with (1+disp)adding 1 to the result each time it is displayed (or at least that clause tells J to keep repeating that operation until it gives the same value back twice in a row - which won't happen - or to stop when the machine stops - like if the power is turned off - or if J is shut down - or...).

We use arbitrary precision numbers, not because there's any likelihood that fixed width numbers would ever overflow, but to emphasize that this thing is going to have to be shut down by some mechanism outside the program.

That said... note that what we are doing here is counting using an internal representation and converting that to octal for display. If we instead wanted to add 1 to a value provided to us in octal and provide the result in octal, we might instead use >: (add 1) and wrap it in &.(8&#.) (convert argument from octal and use inverse on result) with a list of digits representing our octal number. For example:

   >:&.(8&#.)7 6
7 7
   >:&.(8&#.)7 7
1 0 0
   >:&.(8&#.)1 0 0
1 0 1

Java[edit]

public class Count{
    public static void main(String[] args){
        for(int i = 0;i >= 0;i++){
            System.out.println(Integer.toOctalString(i)); //optionally use "Integer.toString(i, 8)"
        }
    }
}

JavaScript[edit]

for (var n = 0; n < 1e14; n++) { // arbitrary limit that's not too big
    document.writeln(n.toString(8)); // not sure what's the best way to output it in JavaScript
}

jq[edit]

Here we use JSON strings of octal digits to represent octal numbers, and therefore there is no language-defined upper bound for the problem. We are careful therefore to select an algorithm that will continue indefinitely so long as there are sufficient physical resources. This is done by framing the problem so that we can use jq's `recurse(_)`.

# generate octals as strings, beginning with "0"
def octals:
  # input and output: array of octal digits in reverse order
  def octal_add1:
    [foreach (.[], null) as $d ({carry: 1};
     if $d then ($d + .carry ) as $r
     | if $r > 7
       then {carry: 1, emit: ($r - 8)}
       else {carry: 0, emit: $r }
       end
     elif (.carry == 0) then .emit = null
     else .emit = .carry
     end;
  select(.emit).emit)];
  
  [0] | recurse(octal_add1) | reverse | join("");

octals

To print the octal strings without quotation marks, invoke jq with the -r command-line option.

Julia[edit]

for i in one(Int64):typemax(Int64)
    print(oct(i), " ")
    sleep(0.1)
end

I slowed the loop down with a sleep to make it possible to see the result without being swamped.

Output:
1 2 3 4 5 6 7 10 11 12 13 14 15 16 17 20 21 22 23 24 25 26 27 30 31 32 33 34 35 36 ^C

Klingphix[edit]

include ..\Utilitys.tlhy

:octal "" >ps [dup 7 band tostr ps> chain >ps 8 / int] [dup abs 0 >] while ps> tonum bor ;

( 0 10 ) sequence @octal map pstack

" " input
Output:
((0, 1, 2, 3, 4, 5, 6, 7, 10, 11, 12))

Kotlin[edit]

//  version 1.1

//  counts up to 177 octal i.e. 127 decimal
fun main(args: Array<String>) {
    (0..Byte.MAX_VALUE).forEach { println("%03o".format(it)) }
}
Output:

First ten lines:

000
001
002
003
004
005
006
007
010
011

LabVIEW[edit]

LabVIEW contains a Number to Octal String function. The following image shows the front panel and block diagram.
Count in octal.png

Lang5[edit]

'%4o '__number_format set
0 do dup 1 compress . "\n" . 1 + loop

langur[edit]

We have to use an arbitrary limit for this.

We use the :8x interpolation modifier to create a string in base 8 (may use base 2 to 36).

val .limit = 70000

for .i = 0; .i <= .limit; .i += 1 {
    writeln $"10x\.i; == 8x\.i:8x;"
}
Output:
10x0 == 8x0
10x1 == 8x1
10x2 == 8x2
10x3 == 8x3
10x4 == 8x4
10x5 == 8x5
10x6 == 8x6
10x7 == 8x7
10x8 == 8x10
10x9 == 8x11
10x10 == 8x12
10x11 == 8x13
10x12 == 8x14
10x13 == 8x15
10x14 == 8x16
10x15 == 8x17
10x16 == 8x20
10x17 == 8x21
10x18 == 8x22
10x19 == 8x23
10x20 == 8x24
10x21 == 8x25
10x22 == 8x26
10x23 == 8x27
10x24 == 8x30
10x25 == 8x31
10x26 == 8x32
10x27 == 8x33
10x28 == 8x34
10x29 == 8x35
10x30 == 8x36
10x31 == 8x37
10x32 == 8x40
10x33 == 8x41
10x34 == 8x42
10x35 == 8x43
10x36 == 8x44
10x37 == 8x45
10x38 == 8x46
10x39 == 8x47
10x40 == 8x50
10x41 == 8x51
10x42 == 8x52
10x43 == 8x53
10x44 == 8x54
10x45 == 8x55
10x46 == 8x56
10x47 == 8x57
10x48 == 8x60
10x49 == 8x61
10x50 == 8x62
10x51 == 8x63
10x52 == 8x64
...
10x69982 == 8x210536
10x69983 == 8x210537
10x69984 == 8x210540
10x69985 == 8x210541
10x69986 == 8x210542
10x69987 == 8x210543
10x69988 == 8x210544
10x69989 == 8x210545
10x69990 == 8x210546
10x69991 == 8x210547
10x69992 == 8x210550
10x69993 == 8x210551
10x69994 == 8x210552
10x69995 == 8x210553
10x69996 == 8x210554
10x69997 == 8x210555
10x69998 == 8x210556
10x69999 == 8x210557
10x70000 == 8x210560

LFE[edit]

(: lists foreach
  (lambda (x)
    (: io format '"~p~n" (list (: erlang integer_to_list x 8))))
  (: lists seq 0 2000))

Liberty BASIC[edit]

Terminate these ( essentially, practically) infinite loops by hitting <CTRL<BRK>

    'the method used here uses the base-conversion from RC Non-decimal radices/Convert
    'to terminate hit <CTRL<BRK>

    global      alphanum$
    alphanum$   ="01234567"

    i =0

    while 1
        print toBase$( 8, i)
        i =i +1
    wend

    end

    function toBase$( base, number) '   Convert decimal variable to number string.
        maxIntegerBitSize   =len( str$( number))
        toBase$             =""
        for i =10 to 1 step -1
            remainder   =number mod base
            toBase$     =mid$( alphanum$, remainder +1, 1) +toBase$
            number      =int( number /base)
            if number <1 then exit for
        next i
        toBase$ =right$( "             " +toBase$, 10)
    end function

As suggested in LOGO, it is easy to work on a string representation too.

 op$ = "00000000000000000000"
L   =len( op$)

while 1
    started =0

    for i =1 to L
        m$ =mid$( op$, i, 1)
        if started =0 and m$ ="0" then print " "; else print m$;: started =1
    next i
    print

    for i =L to 1 step -1
        p$ =mid$( op$, i, 1)
        if p$ =" " then v =0 else v =val( p$)
        incDigit  = v +carry
        if i =L then incDigit =incDigit +1
        if incDigit >=8 then
            replDigit =incDigit -8
            carry     =1
        else
            replDigit =incDigit
            carry     =0
        end if
        op$ =left$( op$, i -1) +chr$( 48 +replDigit) +right$( op$, L -i)
    next i

wend

end

Or use a recursive listing of permutations with the exception that the first digit is not 0 (unless listing single-digit numbers). For each digit-place, list numbers with 0-7 in the next digit-place.

 i = 0
while 1
    call CountOctal 0, i, i > 0
    i = i + 1
wend

sub CountOctal value, depth, startValue
    value = value * 10
    for i = startValue to 7
        if depth > 0 then
            call CountOctal value + i, depth - 1, 0
        else
            print value + i
        end if
    next i
end sub

[edit]

No built-in octal-formatting, so it's probably more efficient to just manually increment a string than to increment a number and then convert the whole thing to octal every time we print. This also lets us keep counting as long as we have room for the string.

to increment_octal :n
  ifelse [empty? :n] [
    output 1
  ] [
    local "last
    make "last last :n
    local "butlast
    make "butlast butlast :n
    make "last sum :last 1
    ifelse [:last < 8] [
      output word :butlast :last
    ] [
      output word (increment_octal :butlast) 0
    ]
  ]
end

make "oct 0
while ["true] [
  print :oct
  make "oct increment_octal :oct
]

LOLCODE[edit]

LOLCODE has no conception of octal numbers, but we can use string concatenation (SMOOSH) and basic arithmetic to accomplish the task.

HAI 1.3

HOW IZ I octal YR num
    I HAS A digit, I HAS A oct ITZ ""
    IM IN YR octalizer
        digit R MOD OF num AN 8
        oct R SMOOSH digit oct MKAY
        num R QUOSHUNT OF num AN 8
        NOT num, O RLY?
            YA RLY, FOUND YR oct
        OIC
    IM OUTTA YR octalizer
IF U SAY SO

IM IN YR printer UPPIN YR num
    VISIBLE I IZ octal YR num MKAY
IM OUTTA YR printer

KTHXBYE

Lua[edit]

for l=1,2147483647 do
  print(string.format("%o",l))
end

M4[edit]

define(`forever',
   `ifelse($#,0,``$0'',
   `pushdef(`$1',$2)$4`'popdef(`$1')$0(`$1',eval($2+$3),$3,`$4')')')dnl
forever(`y',0,1, `eval(y,8)
')

Maple[edit]

octcount := proc (n)
 seq(printf("%a \n", convert(i, octal)), i = 1 .. n);
 end proc;

MACRO-10[edit]

title  OCTAL - Count in octal.
subttl PDP-10 assembly (MACRO-10 on TOPS-20). KJX 2022.
search monsym,macsym

comment \
        If you want to see the overflow happening without waiting for
        too long, change "movei b,1" to "move b,[377777,,777770]".
        \

a=:1                                            ;Names for accumulators.
b=:2
c=:3

define crlf <tmsg <
>>                                              ;Macro to print newline.

start:: reset%                                  ;Initialize process.
        movei b,1                               ;B is the counter.
        movei c,^d8                             ;Octal output (nout%).
        do.
           movei a,.priou                       ;Use standard-output (nout%).
           nout%                                ;Print number in B.
              jrst [ tmsg <Output error.>       ;   NOUT can fail, print err-msg
                     jrst endprg ]              ;   and stop in that case.
           crlf                                 ;Print newline.
           aos b                                ;Add one to B.
           jfcl 10,[ tmsg <Arithmetic Overflow (AROV).> ;Handle overflow.
                     jrst endprg ]
           loop.                                ;Do again.
        enddo.

endprg: haltf%                                  ;Halt program.
        jrst start                              ;Allow continue-command.

end start

Mathematica/Wolfram Language[edit]

x=0;
While[True,Print[BaseForm[x,8];x++]

MATLAB / Octave[edit]

    n = 0; 
    while (1)
        dec2base(n,8)
        n = n+1; 
    end;

Or use printf:

    n = 0; 
    while (1)
        printf('%o\n',n);
        n = n+1; 
    end;

If a predefined sequence should be displayed, one can use

    seq = 1:100;
    dec2base(seq,8)

or

    printf('%o\n',seq);

Mercury[edit]

:- module count_in_octal.
:- interface.
:- import_module io.

:- pred main(io::di, io::uo) is det.

:- implementation.
:- import_module int, list, string.

main(!IO) :-
    count_in_octal(0, !IO).

:- pred count_in_octal(int::in, io::di, io::uo) is det.

count_in_octal(N, !IO) :-
    io.format("%o\n", [i(N)], !IO),
    count_in_octal(N + 1, !IO).

min[edit]

Works with: min version 0.19.3

min has no support for octal or base conversion (it is a minimalistic language, after all) so we need to do that ourselves.

(
  (dup 0 ==) (pop () 0 shorten)
  (((8 mod) (8 div)) cleave) 'cons linrec
  reverse 'print! foreach newline
) :octal

0 (dup octal succ)
9.223e18 int times ; close to max int value

МК-61/52[edit]

ИП0	П1	1	0	/	[x]	П1	Вx	{x}	1
0	*	7	-	x=0	21	ИП1	x#0	28	БП
02	ИП0	1	+	П0	С/П	БП	00	ИП0	lg
[x]	1	+	10^x	П0	С/П	БП	00

Modula-2[edit]

MODULE octal;

IMPORT  InOut;

VAR     num             : CARDINAL;

BEGIN
  num := 0;
  REPEAT
    InOut.WriteOct (num, 12);           InOut.WriteLn;
    INC (num)
  UNTIL num = 0
END octal.

Nanoquery[edit]

Translation of: C

Even though all integers are arbitrary precision, the maximum value that can be represented as octal using the format function is 2^64 - 1. Once this value is reached, the program terminates.

i = 0
while i < 2^64
    println format("%o", i)
    i += 1
end

NetRexx[edit]

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

import java.math.BigInteger

-- allow an option to change the output radix.
parse arg radix .
if radix.length() == 0 then radix = 8
k_ = BigInteger
k_ = BigInteger.ZERO

loop forever
  say k_.toString(int radix)
  k_ = k_.add(BigInteger.ONE)
  end

NewLISP[edit]

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

; Although NewLISP itself uses a 64-bit integer representation, the
; format function relies on underlying C library's printf function,
; which can only handle a 32-bit octal number on this implementation.

(for (i 0 (pow 2 32)) (println (format "%o" i)))

(exit)

Sample output:

0
1
2
3
4
5
6
7
10
11
12
...

Nim[edit]

import strutils
for i in 0 ..< int.high:
  echo toOct(i, 16)

Oberon-2[edit]

Works with: oo2c
MODULE CountInOctal;
IMPORT
  NPCT:Tools,
  Out := NPCT:Console;
VAR
  i: INTEGER;

BEGIN
  FOR i := 0 TO MAX(INTEGER) DO;
    Out.String(Tools.IntToOct(i));Out.Ln
  END
END CountInOctal.
Output:
00000000000
00000000001
00000000002
00000000003
00000000004
00000000005
00000000006
00000000007
00000000010
00000000011
00000000012
00000000013
00000000014
00000000015
00000000016
00000000017
00000000020
00000000021
...
00000077757
00000077760
00000077761
00000077762
00000077763
00000077764
00000077765
00000077766
00000077767
00000077770
00000077771
00000077772
00000077773
00000077774
00000077775
00000077776
00000077777

OCaml[edit]

let () =
  for i = 0 to max_int do
    Printf.printf "%o\n" i
  done
Output:
0
1
2
3
4
5
6
7
10
11
12
...
7777777775
7777777776
7777777777

PARI/GP[edit]

Both versions will count essentially forever; the universe will succumb to proton decay long before the counter rolls over even in the 32-bit version.

Manual:

oct(n)=n=binary(n);if(#n%3,n=concat([[0,0],[0]][#n%3],n));forstep(i=1,#n,3,print1(4*n[i]+2*n[i+1]+n[i+2]));print;
n=0;while(1,oct(n);n++)

Automatic:

Works with: PARI/GP version 2.4.3 and above
n=0;while(1,printf("%o\n",n);n++)

Pascal[edit]

See Delphi or
Works with: Free Pascal

old string incrementer for Turbo Pascal transformed, same as in http://rosettacode.org/wiki/Count_in_octal#Logo, about 100x times faster than Dephi-Version, with the abilty to used preformated strings leading zeroes. Added a Bit fiddling Version IntToOctString, nearly as fast.

program StrAdd;
{$Mode Delphi}
{$Optimization ON}
uses
  sysutils;//IntToStr

const
  maxCntOct = (SizeOf(NativeUint)*8+(3-1)) DIV 3;

procedure IntToOctString(i: NativeUint;var res:Ansistring);
var
  p : array[0..maxCntOct] of byte;
  c,cnt: LongInt;
begin
  cnt := maxCntOct;
  repeat
    c := i AND 7;
    p[cnt] := (c+Ord('0'));
    dec(cnt);
    i := i shr 3;
  until (i = 0);
  i := cnt+1;
  cnt := maxCntOct-cnt;
  //most time consuming with Ansistring
  //call fpc_ansistr_unique
  setlength(res,cnt);
  move(p[i],res[1],cnt);
end;

procedure IncStr(var s:String;base:NativeInt);
var
  le,c,dg:nativeInt;
begin
  le := length(s);
  IF le = 0 then
  Begin
    s := '1';
    EXIT;
  end;

  repeat
    dg := ord(s[le])-ord('0') +1;
    c  := ord(dg>=base);
    dg := dg-(base AND (-c));
    s[le] := chr(dg+ord('0'));
    dec(le);
  until (c = 0) or (le<=0);

  if (c = 1) then
  begin
    le := length(s);
    setlength(s,le+1);
    move(s[1],s[2],le);
    s[1] := '1';
  end;
end;

const
  MAX = 8*8*8*8*8*8*8*8*8;//8^9
var
  sOct,
  s  : AnsiString;
  i : nativeInt;
  T1,T0: TDateTime;
Begin
  sOct := '';
  For i := 1 to 16 do
  Begin
    IncStr(sOct,8);
    writeln(i:10,sOct:10);
  end;
  writeln;

  For i := 1 to 16 do
  Begin
    IntToOctString(i,s);
    writeln(i:10,s:10);
  end;

  sOct := '';
  T0 := time;
  For i := 1 to MAX do
    IncStr(sOct,8);
  T0 := (time-T0)*86400;
  writeln(sOct);

  T1 := time;
  For i := 1 to MAX do
    IntToOctString(i,s);
  T1 := (time-T1)*86400;
  writeln(s);
  writeln;
  writeln(MAX);
  writeln('IncStr         ',T0:8:3);
  writeln('IntToOctString ',T1:8:3);
end.
Output:
         1         1
         2         2
         3         3
         4         4
         5         5
         6         6
         7         7
         8        10
         9        11
        10        12
        11        13
        12        14
        13        15
        14        16
        15        17
        16        20

         1         1
         2         2
         3         3
         4         4
         5         5
         6         6
         7         7
         8        10
         9        11
        10        12
        11        13
        12        14
        13        15
        14        16
        15        17
        16        20

1000000000
1000000000

134217728
IncStr            0.944 secs
IntToOctString    2.218 secs

A recursive approach[edit]

For this task, recursion offers no advantage in clarity or efficiency over iteration, but it's nevertheless an instructive exercise.

program OctalCount;

{$mode objfpc}{$H+}

var
  i : integer;

// display n in octal on console
procedure PutOctal(n : integer);
var
  digit, n3 : integer;
begin
  n3 := n shr 3;
  if n3 <> 0 then PutOctal(n3);
  digit := n and 7;
  write(digit);
end;

// count in octal until integer overflow
begin
  i := 1;
  while i > 0 do
    begin
       PutOctal(i);
       writeln;
       i := i + 1;
    end;
  readln;
end.
Output:

Showing last 10 lines of output

17777777766
17777777767
17777777770
17777777771
17777777772
17777777773
17777777774
17777777775
17777777776
17777777777

Perl[edit]

Since task says "system register", I take it to mean "no larger than machine native integer limit":

use POSIX;
printf "%o\n", $_ for (0 .. POSIX::UINT_MAX);

Otherwise:

use bigint;
my $i = 0;
printf "%o\n", $i++ while 1

The above count in binary or decimal and convert to octal. This actually counts in octal. It will run forever or until the universe ends, whichever comes first.

#!/usr/bin/perl

$_ = 0;
s/([^7])?(7*)$/ $1 + 1 . $2 =~ tr!7!0!r /e while print "$_\n";

Phix[edit]

without javascript_semantics
integer i = 0
constant ESC = #1B
while not find(get_key(),{ESC,'q','Q'}) do
    printf(1,"%o\n",i)
    i += 1
end while

See Integer_sequence#Phix for something that will run in a browser, obviously use "%o" instead of "%d" to make it display octal numbers, or more accurately in that case, mpz_get_str(i,8).

PHP[edit]

<?php
for ($n = 0; is_int($n); $n++) {
  echo decoct($n), "\n";
}
?>

Picat[edit]

Ways to convert to octal numbers:

  • to_oct_string(N)
  • to_radix_string(N,8)
  • printf("%o\n",N)


go =>
  gen(N),
  println(to_oct_string(N)),
  fail.

gen(I) :-
  gen(0, I).
gen(I, I).
gen(I, J) :-
  I2 is I + 1,
  gen(I2, J).
Output:
0
1
2
3
4
5
6
7
10
11
...
17615737040105402212262317777777776
17615737040105402212262317777777777
17615737040105402212262320000000000
17615737040105402212262320000000001
17615737040105402212262320000000002
<Ctrl-C>


PicoLisp[edit]

(for (N 0  T  (inc N))
   (prinl (oct N)) )

Pike[edit]

int i=1;
while(true)
    write("0%o\n", i++);
Output:
01
02
...

PL/I[edit]

Version 1:

/* Do the actual counting in octal. */
count: procedure options (main);
   declare v(5) fixed(1) static initial ((5)0);
   declare (i, k) fixed;

   do k = 1 to 999;
      call inc;
      put skip edit ( (v(i) do i = 1 to 5) ) (f(1));
   end;

inc: proc;
   declare (carry, i) fixed binary;

   carry = 1;
   do i = 5 to 1 by -1;
      v(i) = v(i) + carry;
      if v(i) > 7 then
         do; v(i) = v(i) - 8; if i = 1 then stop; carry = 1; end;
      else
         carry = 0;
   end;
end inc;

end count;

Version 2:

count: procedure options (main); /* 12 Jan. 2014 */
   declare (i, j) fixed binary;

   do i = 0 upthru 2147483647;
      do j = 30 to 0 by -3;
         put edit (iand(isrl(i, j), 7) ) (f(1));
      end;
      put skip;
   end;

end count;
Output:
(End of) Output of version 1
00000001173
00000001174
00000001175
00000001176
00000001177
00000001200
00000001201
00000001202
00000001203
00000001204
00000001205
00000001206
00000001207
00000001210
00000001211
00000001212
00000001213
00000001214
00000001215
00000001216

PL/I-80[edit]

If you only need to count, and aren't bothered by leading zeroes in the output, this will do the trick simply and with a minimum of fuss.

octal_count:
    procedure options (main);
    dcl i fixed;
    i = 1;
    do while (i ^= 0);
      put skip edit (unspec(i)) (b3);
      i = i + 1;
    end;

end octal_count;
Output:

First and last 10 numbers of output

000001
000002
000003
000004
000005
000006
000007
000010
000011
000012
  ...
177766
177767
177770
177771
177772
177773
177774
177775
177776
177777
But a general purpose function to return the octal representation of an integer value as a string (similar to the OCT$ function in many BASICs) may prove more useful.
octal_count:
    procedure options (main);
    dcl i fixed;
    i = 1;
    do while (i ^=  0);
       put skip list (octal(i));
       i = i + 1;
    end;

    stop;

octal:
    procedure (n) returns (char(6) varying);
    dcl
       (n, m) fixed,
       s char(6) varying;
    /* n is passed by reference, so make a local copy */
    m = n; 
    s = '';
    do while (m > 0);
       s = ascii(mod(m,8) + rank('0')) || s;
       m = m / 8;
    end;
    return (s);
end octal;

end octal_count;
Output:

First and last 10 numbers of output

1
2
3
4
5
6
7
10
11
12
 ...
77766
77767
77770
77771
77772
77773
77774
77775
77776
77777

PL/M[edit]

100H: /* PRINT INTEGERS IN OCTAL */
   BDOS: PROCEDURE( FN, ARG ); /* CP/M BDOS SYSTEM CALL */
      DECLARE FN BYTE, ARG ADDRESS;
      GOTO 5;
   END BDOS;
   PR$CHAR:  PROCEDURE( C ); DECLARE C BYTE; CALL BDOS( 2, C ); END;
   PR$NL:    PROCEDURE; CALL PR$CHAR( 0DH ); CALL PR$CHAR( 0AH ); END;
   PR$OCTAL: PROCEDURE( N );
      DECLARE N ADDRESS;
      DECLARE V ADDRESS, O$STR( 7 ) BYTE, W BYTE;
      V = N;
      O$STR( W := 0 ) = '0' + ( V AND 7 );
      DO WHILE( ( V := SHR( V, 3 ) ) > 0 );
         O$STR( W := W + 1 ) = '0' + ( V AND 7 );
      END;
      W = W + 1;
      DO WHILE( W <> 0 );
         CALL PR$CHAR( O$STR( W := W - 1 ) );
      END;
   END PR$OCTAL;

   DECLARE N ADDRESS;
   N = 0;
   CALL PR$OCTAL( N );
   CALL PR$NL;
   DO WHILE( ( N := N + 1 ) > 0 ); /* AFTER 65535 N WILL WRAP 'ROUND TO 0 */
      CALL PR$OCTAL( N );
      CALL PR$NL;
   END;
EOF

PowerShell[edit]

[int64]$i = 0
While ( $True )
    {
    [Convert]::ToString( ++$i, 8 )
    }

Prolog[edit]

Rather than just printing out a list of octal numbers, this code will generate a sequence. octal/1 can also be used to tell if a number is a valid octal number or not. octalize will keep producing and printing octal number, there is no limit.

o(O) :- member(O, [0,1,2,3,4,5,6,7]).

octal([O]) :- o(O).
octal([A|B]) :- 
	octal(O), 
	o(T), 
	append(O, [T], [A|B]),
	dif(A, 0).
	
octalize :-
	forall(
		octal(X),
		(maplist(write, X), nl)
	).

PureBasic[edit]

Procedure.s octal(n.q)
  Static Dim digits(20)
  Protected i, j, result.s
  For i = 0 To 20
    digits(i) = n % 8
    n / 8
    If n < 1
      For j = i To 0 Step -1
        result + Str(digits(j))
      Next 
      Break
    EndIf
  Next 
  
  ProcedureReturn result  
EndProcedure

Define n.q
If OpenConsole()
  While n >= 0
    PrintN(octal(n))
    n + 1
  Wend 
  
  Print(#CRLF$ + #CRLF$ + "Press ENTER to exit"): Input()
  CloseConsole()
EndIf

Sample output:

0
1
2
3
4
5
6
7
10
11
12
...
777777777777777777767
777777777777777777770
777777777777777777771
777777777777777777772
777777777777777777773
777777777777777777774
777777777777777777775
777777777777777777776
777777777777777777777

Python[edit]

import sys
for n in xrange(sys.maxint):
    print oct(n)

QB64[edit]

Dim As Integer iNum, Icount
Dim sMax As String

sMax = ""
Do While Val(sMax) <= 0 Or Val(sMax) > 32767
    Input "Please type a value from 1 to 32767 ", sMax
Loop
iNum = Val(sMax)
For Icount = 0 To iNum Step 1
    Print Oct$(Icount)
Next
End

REM also QBasic example runs under QB64

Quackery[edit]

8 base put
0 [ dup echo cr 1+ again ]

Racket[edit]

#lang racket
(for ([i (in-naturals)])
  (displayln (number->string i 8)))

(Racket has bignums, so this loop will never end.)

Raku[edit]

(formerly Perl 6)

say .base(8) for ^Inf;
Output:
0

Here we arbitrarily show as many lines of output as there are lines in the program. :-)

REXX[edit]

If this REXX program wouldn't be stopped, it would count forever.

The technique used is to convert the decimal number to binary, and separate the binary digits in groups of three, and then convert those binary groups (numbers) to decimal.

/*REXX program counts in octal until the number exceeds the number of program statements*/

        /*┌────────────────────────────────────────────────────────────────────┐
          │ Count all the protons  (and electrons!)  in the universe.          │
          │                                                                    │
          │ According to Sir Arthur Eddington in 1938 at his Tamer Lecture at  │
          │ Trinity College (Cambridge),  he postulated that there are exactly │
          │                                                                    │
          │                              136 ∙ 2^256                           │
          │                                                                    │
          │ protons in the universe,  and the same number of electrons,  which │
          │ is equal to around  1.57477e+79.                                   │
          │                                                                    │
          │ [Although, a modern estimate is around  10^80.]                    │
          └────────────────────────────────────────────────────────────────────┘*/

numeric digits 100000                            /*handle almost any sized big numbers. */
numIn= right('number in', 20)                    /*used for indentation of the output.  */
w= length( sourceline() )                        /*used for formatting width of numbers.*/

  do #=0  to 136 * (2**256)                      /*Sir Eddington, here we come !        */
  != x2b( d2x(#) )
  _= right(!,  3 * (length(_) % 3 + 1),  0)
  o=
                do k=1  to length(_)  by 3
                o= o'0'substr(_, k, 3)
                end   /*k*/

  say numIn  'base ten = '   right(#,w) numIn    "octal = "   right( b2x(o) + 0, w + w)
  if #>sourceline()  then leave                  /*stop if # of protons > pgm statements*/
  end   /*#*/
                                                 /*stick a fork in it,  we're all done. */
output:
           number in base ten =   0            number in octal =     0
           number in base ten =   1            number in octal =     1
           number in base ten =   2            number in octal =     2
           number in base ten =   3            number in octal =     3
           number in base ten =   4            number in octal =     4
           number in base ten =   5            number in octal =     5
           number in base ten =   6            number in octal =     6
           number in base ten =   7            number in octal =     7
           number in base ten =   8            number in octal =    10
           number in base ten =   9            number in octal =    11
           number in base ten =  10            number in octal =    12
           number in base ten =  11            number in octal =    13
           number in base ten =  12            number in octal =    14
           number in base ten =  13            number in octal =    15
           number in base ten =  14            number in octal =    16
           number in base ten =  15            number in octal =    17
           number in base ten =  16            number in octal =    20
           number in base ten =  17            number in octal =    21
           number in base ten =  18            number in octal =    22
           number in base ten =  19            number in octal =    23
           number in base ten =  20            number in octal =    24
           number in base ten =  21            number in octal =    25
           number in base ten =  22            number in octal =    26
           number in base ten =  23            number in octal =    27
           number in base ten =  24            number in octal =    30
           number in base ten =  25            number in octal =    31
           number in base ten =  26            number in octal =    32
           number in base ten =  27            number in octal =    33
           number in base ten =  28            number in octal =    34
           number in base ten =  29            number in octal =    35
           number in base ten =  30            number in octal =    36
           number in base ten =  31            number in octal =    37
           number in base ten =  32            number in octal =    40
           number in base ten =  33            number in octal =    41 

Ring[edit]

size = 30
for n = 1 to size
    see octal(n) + nl
next

func octal m
     output = ""
     w = m
     while fabs(w) > 0    
           oct = w & 7
           w = floor(w / 8)
           output = string(oct) + output
     end
     return output

Ruby[edit]

n = 0
loop do
  puts "%o" % n
  n += 1
end

# or
for n in (0..)
  puts n.to_s(8)
end

# or
0.upto(1/0.0) do |n|
  printf "%o\n", n
end

# version 2.1 later
0.step do |n|
  puts format("%o", n)
end

Run BASIC[edit]

input "Begin number:";b
input "  End number:";e
 
for i = b to e
  print i;" ";toBase$(8,i)
next i 
end
 
function toBase$(base,base10)
for i = 10 to 1 step -1
  toBase$   = str$(base10 mod base) +toBase$
  base10    = int(base10 / base)
  if base10 < 1 then exit for
next i
end function

Rust[edit]

fn main() {
    for i in 0..std::usize::MAX {
        println!("{:o}", i);
    }
}

Salmon[edit]

Salmon has built-in unlimited-precision integer arithmetic, so these examples will all continue printing octal values indefinitely, limited only by the amount of memory available (it requires O(log(n)) bits to store an integer n, so if your computer has 1 GB of memory, it will count to a number with on the order of octal digits).

iterate (i; [0...+oo])
    printf("%o%\n", i);;

or

for (i; 0; true)
    printf("%o%\n", i);;

or

variable i := 0;
while (true)
  {
    printf("%o%\n", i);
    ++i;
  };

S-BASIC[edit]

Although many BASICs have a built-in OCT$ function, S-BASIC does not, so we have to supply our own

rem - return p mod q
function mod(p, q = integer) = integer
end = p - q * (p / q)

rem - return octal representation of n
function oct$(n = integer) = string
var s = string
s = ""
while n > 0 do
  begin
     s = chr(mod(n,8) + '0') + s
     n = n / 8
  end
end = s

rem - count in octal until overflow
var i = integer
i = 1
while i > 0 do
  begin
    print oct$(i)
    i = i + 1
  end

end
Output:

Showing first and last 10 lines of output

1
2
3
4
5
6
7
10
11
12
...
77766
77767
77770
77771
77772
77773
77774
77775
77776
77777

Scala[edit]

Stream from 0 foreach (i => println(i.toOctalString))

Scheme[edit]

(do ((i 0 (+ i 1))) (#f) (display (number->string i 8)) (newline))

Scratch[edit]

ScratchCountInOctal.png

Seed7[edit]

This example uses the radix operator to write a number in octal.

$ include "seed7_05.s7i";
 
const proc: main is func
  local
    var integer: i is 0;
  begin
    repeat
      writeln(i radix 8);
      incr(i);
    until FALSE;
  end func;

Sidef[edit]

var i = 0;
loop { say i++.as_oct }

Simula[edit]

BEGIN

    PROCEDURE OUTOCT(N); INTEGER N;
    BEGIN
        PROCEDURE OCT(N); INTEGER N;
        BEGIN
            IF N > 0 THEN BEGIN
                OCT(N//8);
                OUTCHAR(CHAR(RANK('0')+MOD(N,8)));
            END;
        END OCT;
        IF N < 0 THEN BEGIN OUTCHAR('-'); OUTOCT(-N); END
        ELSE IF N = 0 THEN OUTCHAR('0')
        ELSE OCT(N);
    END OUTOCT;

    INTEGER I;
    WHILE I < MAXINT DO BEGIN
        OUTINT(I,0);
        OUTTEXT(" => ");
        OUTOCT(I);
        OUTIMAGE;
        I := I+1;
    END;
END.

Smalltalk[edit]

Works with: Smalltalk/X
0 to:Integer infinity do:[:n |
    n printOn:Stdout radix:8.
    Stdout cr.
]
Output:
1
2
3
4
5
6
7
10
11
12
13
14
15
16
17
20
21
...

Sparkling[edit]

for (var i = 0; true; i++) {
    printf("%o\n", i);
}

Standard ML[edit]

local
  fun count n = (print (Int.fmt StringCvt.OCT n ^ "\n"); count (n+1))
in
  val _ = count 0
end

Swift[edit]

import Foundation

func octalSuccessor(value: String) -> String {
   if value.isEmpty {
        return "1"
   } else {
     let i = value.startIndex, j = value.endIndex.predecessor()
     switch (value[j]) {
       case "0": return value[i..<j] + "1"
       case "1": return value[i..<j] + "2"
       case "2": return value[i..<j] + "3"
       case "3": return value[i..<j] + "4"
       case "4": return value[i..<j] + "5"
       case "5": return value[i..<j] + "6"
       case "6": return value[i..<j] + "7"
       case "7": return octalSuccessor(value[i..<j]) + "0"
       default:
         NSException(name:"InvalidDigit", reason: "InvalidOctalDigit", userInfo: nil).raise();
         return ""
     }
  }
}

var n = "0"
while strtoul(n, nil, 8) < UInt.max {
  println(n)
  n = octalSuccessor(n)
}
Output:

The first few lines. anyway:

0
1
2
3
4
5
6
7
10
11
12
13
14
15
16
17
20
21
22
23

Tcl[edit]

package require Tcl 8.5;   # arbitrary precision integers; we can count until we run out of memory!
while 1 {
    puts [format "%llo" [incr counter]]
}

UNIX Shell[edit]

We use the bc calculator to increment our octal counter:

#!/bin/sh
num=0
while true; do
  echo $num
  num=`echo "obase=8;ibase=8;$num+1"|bc`
done

Using printf[edit]

Increment a decimal counter and use printf(1) to print it in octal. Our loop stops when the counter overflows to negative.

#!/bin/sh
num=0
while test 0 -le $num; do
  printf '%o\n' $num
  num=`expr $num + 1`
done

Various recent shells have a bultin $(( ... )) for arithmetic rather than running expr, in which case

Works with: bash
Works with: pdksh version 5.2.14
num=0
while test 0 -le $num; do
  printf '%o\n' $num
  num=$((num + 1))
done

VBA[edit]

With i defined as an Integer, the loop will count to 77777 (32767 decimal). Error handling added to terminate nicely on integer overflow.

Sub CountOctal()
Dim i As Integer
i = 0
On Error GoTo OctEnd
Do
    Debug.Print Oct(i)
    i = i + 1
Loop
OctEnd:
Debug.Print "Integer overflow - count terminated"
End Sub

VBScript[edit]

For i = 0 To 20
	WScript.StdOut.WriteLine Oct(i)
Next

Vim Script[edit]

let counter = 0
while counter >= 0
    echon printf("%o\n", counter)
    let counter += 1
endwhile

Vlang[edit]

import math
fn main() {
    for i := i8(0); ; i++ {
        println("${i:o}")
        if i == math.max_i8 {
            break
        }
    }
}
Output:
0
1
2
...
173
174
175
176
177

VTL-2[edit]

Stops at 65535, the largest integer supported by VTL-2.

1000 N=0
1010 #=2000
1020 ?=""
1030 #=N=65535*9999
1040 N=N+1
1050 #=1010
2000 R=!
2010 O=N
2020 D=1
2030 O=O/8
2040 :D)=%
2050 D=D+1
2060 #=O>1*2030
2070 E=D-1
2080 $=48+:E)
2090 E=E-1
2100 #=E>1*2080
2110 #=R
Output:
0
1
2
3
4
5
6
7
10
11
12
...
177775
177776
177777

Whitespace[edit]

This program prints octal numbers until the internal representation of the current integer overflows to -1; it will never do so on some interpreters.

It was generated from the following pseudo-Assembly.

push 0
; Increment indefinitely.
0:
    push -1 ; Sentinel value so the printer knows when to stop.
    copy 1
    call 1
    push 10
    ochr
    push 1
    add
    jump 0
; Get the octal digits on the stack in reverse order.
1:
    dup
    push 8
    mod
    swap
    push 8
    div
    push 0
    copy 1
    sub
    jn 1
    pop
; Print them.
2:
    dup
    jn 3 ; Stop at the sentinel.
    onum
    jump 2
3:
    pop
    ret

Wren[edit]

Library: Wren-fmt
import "/fmt" for Conv

var i = 0
while (true) {
    System.print(Conv.oct(i))
    i = i + 1
}
Output:
0
1
2
3
4
5
6
7
10
11
12
13
14
15
16
17
20
21
22
23
24
^C

XPL0[edit]

XPL0 doesn't have built-in routines to handle octal; instead it uses hex.

include c:\cxpl\codes;     \intrinsic code declarations

proc OctOut(N);    \Output N in octal
int N;
int R;
[R:= N&7;
N:= N>>3;
if N then OctOut(N);
ChOut(0, R+^0);
];

int I;
[I:= 0;
repeat  OctOut(I); CrLf(0);
        I:= I+1;
until KeyHit or I=0;
]

Example output:

0
1
2
3
4
5
6
7
10
11
12
13
14
15
16
17
20
21

zig[edit]

const std = @import("std");
const fmt = std.fmt;
const warn = std.debug.warn;

pub fn main() void {
    var i: u8 = 0;
    var buf: [3]u8 = undefined;

    while (i < 255) : (i += 1) {
        _ = fmt.formatIntBuf(buf[0..], i, 8, false, 0); // buffer, value, base, uppercase, width
        warn("{}\n", buf);
    }
}

Z80 Assembly[edit]

The Sega Master System's screen isn't big enough to show each number on its own line and have all the numbers be visible at the same time, so this program shows 8 per line. Hardware-specific code for loading system font, setting up video display processor, printing, etc. are omitted. Z80 Assembly doesn't have built-in support for displaying octal (or any value in any base for that matter) so it has to be done with a custom routine.

Outputs octal values 0 through 77 (decimal 0 to 63, or hexadecimal 0x00 to 0x3F.)

	xor a          ;LD A,0
	ld b,&40       ;how many numbers to print.
loop_showOctal:
	push af
		push af
			call ShowOctal
			ld a,' '
			call PrintChar  ;put a blank space after the value
		pop af
                
;;;;;;;;;;;;;;;;;;;;; this code starts a new line after every 8th output.
		ld a,b
		and &07 
		dec a             
		call z,NewLine  
;;;;;;;;;;;;;;;;;;;;;; 
	pop af
	inc a  ;next number
	djnz loop_showOctal

	jp $   ;end program

ShowOctal:
	push bc
		ld c,a
		add a
		push af
			ld a,7
			and c
			ld c,a
		pop af
		and &F0
		or c
		and &7F
	pop bc
	jp ShowHex

ShowHex:	 ;this isn't directly below ShowOctal, it's somewhere else entirely.
                 ;thanks to Keith of Chibiakumas for this routine!
	push af
		and %11110000
		ifdef gbz80
			swap a     ;game boy can use this, but Zilog Z80 cannot.
		else
			rrca
			rrca
			rrca
			rrca
		endif
		call PrintHexChar
	pop af
	and %00001111
        ;execution flows into the subroutine below, effectively calling it for us without having to actually do so.
PrintHexChar:
	or a	       ;Clear Carry Flag
	daa
	add a,&F0
	adc a,&40      ;this sequence of instructions converts hexadecimal values to ASCII.
	jp PrintChar   ;hardware-specific routine, omitted. Thanks to Keith of Chibiakumas for this one!
Output:
00 01 02 03 04 05 06 07
10 11 12 13 14 15 16 17
20 21 22 23 24 25 26 27
30 31 32 33 34 35 36 37
40 41 42 43 44 45 46 47
50 51 52 53 54 55 56 57
60 61 62 63 64 65 66 67
70 71 72 73 74 75 76 77

zkl[edit]

foreach n in ([0..]){println("%.8B".fmt(n))}
Output:
0
1
2
3
4
5
6
7
10
11
12

ZX Spectrum Basic[edit]

10 PRINT "DEC.  OCT."
20 FOR i=0 TO 20
30 LET o$="": LET n=i
40 LET o$=STR$ FN m(n,8)+o$
50 LET n=INT (n/8)
60 IF n>0 THEN GO TO 40
70 PRINT i;TAB 3;" = ";o$
80 NEXT i
90 STOP 
100 DEF FN m(a,b)=a-INT (a/b)*b