Count in octal

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

Integer sequence is a similar task without the use of octal numbers.

360 Assembly

The program uses one ASSIST macro (XPRNT) to keep the code as short as possible. <lang 360asm>* 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</lang>

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

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. <lang 6502asm> 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</lang>

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

This assumes the CP/M operating system. The count will terminate after the largest unsigned 16-bit value is reached. <lang> ;------------------------------------------------------- ; 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 ; back to the ccp w/o warm booting ;------------------------------------------------------ ; Console output routine ; print character in A register to console ;------------------------------------------------------ putchr: push h push d push b mov e,a ; character to E for CP/M mvi c,2 ; console output function call bdos ; call on BDOS to perform pop b pop d pop h ret ;------------------------------------------------------ ; output CRLF to console ;------------------------------------------------------ crlf: mvi a,cr call putchr mvi a,lf call putchr ret ;------------------------------------------------------ ; 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 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 shift of 16-bit value in HL right 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 ;------------------------------------------------------- ; data area ;------------------------------------------------------- oldstk: dw 1 stack equ $+128 ; 64 level stack ; end </lang>

Output:

Showing the last 10 lines of the output.

AArch64 Assembly

Works with: as version Raspberry Pi 3B version Buster 64 bits

<lang AArch64 Assembly> /* 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" </lang>

Action!

<lang Action!>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</lang>

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

<lang Ada>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;</lang> First few lines of Output:

Aime

<lang aime>integer o;

o = 0; do {

   o_xinteger(8, o);
   o_byte('\n');
   o += 1;

} while (0 < o);</lang>

ALGOL 68

Works with: ALGOL 68G version Any - tested with release 1.18.0-9h.tiny.

<lang algol68>#!/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

)</lang> Output:

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

ALGOL W

Algol W has built-in hexadecimal and decimal output, this implements octal output. <lang algolw>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.</lang>

Output:

APL

Works with Dyalog APL. 100,000 is just an arbitrarily large number I chose. <lang APL>10⊥¨8∘⊥⍣¯1¨⍳100000</lang>

ARM Assembly

Works with: as version Raspberry Pi

<lang ARM Assembly> /* 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

</lang>

Arturo

<lang rebol>loop 1..40 'i -> print ["number in base 10:" pad to :string i 2 "number in octal:" pad as.octal i 2]</lang>

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

<lang AHK>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 }</lang>

AWK

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:

<lang awk>BEGIN {

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

}</lang>

BASIC

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

Works with: QBasic

<lang qbasic>DIM n AS LONG FOR n = 0 TO &h7FFFFFFF

   PRINT OCT$(n)

NEXT</lang>

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

Works with: Chipmunk Basic

<lang qbasic>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</lang>

Applesoft BASIC

<lang ApplesoftBasic>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</lang>

BASIC256

<lang BASIC256> valor = 0 do print ToOctal(valor) valor += 1 until valor = 0 end </lang>

Commodore BASIC

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.

<lang gwbasic>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</lang>

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

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. <lang basic> 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</lang>

Batch File

<lang dos> @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 </lang>

Output:

BBC BASIC

Terminate by pressing ESCape. <lang bbcbasic> 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$

</lang>

bc

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

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

BCPL

This will count up from 0 until the limit of the machine word. <lang bcpl>get "libhdr"

let start() be $( let x = 0

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

$)</lang>

Befunge

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

Bracmat

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

 ( oct
 =   
   .     !arg:<8
       & (!arg:~<0|ERROR)
     | str$(oct$(div$(!arg.8)) mod$(!arg.8))
 )

& -1:?n & whl'(1+!n:?n&out$(!n oct$!n)); </lang>

Brainf***

<lang bf>+[ 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</lang>

C

<lang C>#include <stdio.h>

int main() {

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

}</lang>

C#

<lang csharp>using System;

class Program {

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

}</lang>

C++

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.

<lang cpp>#include <iostream>

int main() {

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

}</lang>

Clojure

<lang clojure>(doseq [i (range)] (println (format "%o" i)))</lang>

COBOL

Translation of: Delphi

Works with: GNU Cobol version 2.0

<lang cobol> >>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.</lang>

CoffeeScript

<lang coffeescript> n = 0

while true

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

</lang>

Common Lisp

<lang lisp>(loop for i from 0 do (format t "~o~%" i))</lang>

Component Pascal

BlackBox Component Builder <lang oberon2> 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.

</lang> Execute: ^Q CountOctal.Do
Output:

Cowgol

<lang cowgol>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;</lang>

Crystal

<lang ruby># version 0.21.1

using unsigned 8 bit integer, range 0 to 255

(0_u8..255_u8).each { |i| puts i.to_s(8) }</lang>

Output:

D

<lang d>void main() {

   import std.stdio;

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

}</lang>

Dc

Named Macro

A simple infinite loop and octal output will do. <lang Dc>8o0[p1+lpx]dspx</lang>

Anonymous Macro

Needs r (swap TOS and NOS): <lang Dc>8 o 0 [ r p 1 + r dx ] dx</lang> Pushing/poping TOS to a named stack can be used instead of swaping: <lang Dc>8 o 0 [ S@ p 1 + L@ dx ] dx</lang>

DCL

<lang DCL>$ i = 0 $ loop: $ write sys$output f$fao( "!OL", i ) $ i = i + 1 $ goto loop</lang>

Output:

00000000000
00000000001
00000000002
...
17777777777
20000000000
20000000001
...
37777777777
00000000000
00000000001
...

Delphi

<lang Delphi>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.</lang>

Elixir

<lang elixir>Stream.iterate(0,&(&1+1)) |> Enum.each(&IO.puts Integer.to_string(&1,8))</lang> or <lang elixir>Stream.unfold(0, fn n ->

 IO.puts Integer.to_string(n,8)
 {n,n+1}

end) |> Stream.run</lang> or <lang elixir>f = fn ff,i -> :io.fwrite "~.8b~n", [i]; ff.(ff, i+1) end f.(f, 0)</lang>

Emacs Lisp

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

<lang lisp>(dotimes (i most-positive-fixnum) ;; starting from 0

 (message "%o" i))</lang>

Erlang

The fun is copied from Integer sequence#Erlang. I changed the display format. <lang Erlang> F = fun(FF, I) -> io:fwrite("~.8B~n", [I]), FF(FF, I + 1) end. </lang> Use like this:

F( F, 0 ).

Euphoria

<lang euphoria>integer i i = 0 while 1 do

   printf(1,"%o\n",i)
   i += 1

end while</lang>

Output:

F#

<lang fsharp>let rec countInOctal num : unit =

 printfn "%o" num
 countInOctal (num + 1)

countInOctal 1</lang>

Factor

<lang factor>USING: kernel math prettyprint ; 0 [ dup .o 1 + t ] loop</lang>

Forth

Using INTS from Integer sequence#Forth <lang forth>: octal ( -- ) 8 base ! ; \ where unavailable

octal ints</lang>

Fortran

Works with: Fortran version 95 and later

<lang fortran>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</lang>

FreeBASIC

<lang freebasic>' 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</lang>

Frink

<lang frink>i = 0 while true {

   println[i -> octal]
   i = i + 1

}</lang>

Futhark

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

<lang Futhark> 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)

</lang>

FutureBasic

<lang futurebasic> include "ConsoleWindow defstr word

dim as short i

for i = &o000000 to &o000031 // 0 to 25 in decimal

  print oct$(i); " in octal ="; i

next </lang>

Output:

000000 in octal = 0
000001 in octal = 1
000002 in octal = 2
000003 in octal = 3
000004 in octal = 4
000005 in octal = 5
000006 in octal = 6
000007 in octal = 7
000010 in octal = 8
000011 in octal = 9
000012 in octal = 10
000013 in octal = 11
000014 in octal = 12
000015 in octal = 13
000016 in octal = 14
000017 in octal = 15
000020 in octal = 16
000021 in octal = 17
000022 in octal = 18
000023 in octal = 19
000024 in octal = 20
000025 in octal = 21
000026 in octal = 22
000027 in octal = 23
000030 in octal = 24
000031 in octal = 25

Go

<lang go>package main

import (

   "fmt"
   "math"

)

func main() {

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

}</lang> Output:

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: <lang go>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
       }
   }

}</lang> Output:

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: <lang go>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
   }

}</lang> 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. <lang go>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)
   }

}</lang> Output:

Groovy

Size-limited solution: <lang groovy>println 'decimal octal' for (def i = 0; i <= Integer.MAX_VALUE; i++) {

   printf ('%7d  %#5o\n', i, i)

}</lang>

Unbounded solution: <lang groovy>println 'decimal octal' for (def i = 0g; true; i += 1g) {

   printf ('%7d  %#5o\n', i, i)

}</lang>

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

<lang haskell>import Numeric (showOct)

main :: IO () main =

 mapM_
   (putStrLn . flip showOct "")
   [1 ..]</lang>

Icon and Unicon

<lang unicon>link convert # To get exbase10 method

procedure main()

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

end</lang>

J

Solution: <lang J> disp=.([smoutput) ' '(-.~":)8&#.inv

  (1+disp)^:_]0x</lang>

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:

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 tha 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 just to emphasize that this thing is going to have to be shut down by some mechanism outside the program.

Java

<lang java>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)"
       }
   }

}</lang>

JavaScript

<lang javascript>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

}</lang>

jq

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(_)`. <lang jq># 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</lang> To print the octal strings without quotation marks, invoke jq with the -r command-line option.

Julia

<lang Julia> for i in one(Int64):typemax(Int64)

   print(oct(i), " ")
   sleep(0.1)

end </lang> 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

<lang Klingphix>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</lang>

Output:

((0, 1, 2, 3, 4, 5, 6, 7, 10, 11, 12))

Kotlin

<lang scala>// 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)) }

}</lang>

Output:

First ten lines:

LabVIEW

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

Lang5

<lang lang5>'%4o '__number_format set 0 do dup 1 compress . "\n" . 1 + loop</lang>

langur

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).

<lang langur>val .limit = 70000

for .i = 0; .i <= .limit; .i += 1 {

   writeln $"10x\.i; == 8x\.i:8x;"

}</lang>

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

<lang lisp>(: lists foreach

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

</lang>

Liberty BASIC

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

   '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

</lang> As suggested in LOGO, it is easy to work on a string representation too. <lang lb>

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 </lang> 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.

<lang lb>
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 </lang>

Logo

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.

<lang logo>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

]</lang>

LOLCODE

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

Lua

<lang lua>for l=1,2147483647 do

 print(string.format("%o",l))

end</lang>

M4

<lang M4>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) ')</lang>

Maple

<lang Maple> octcount := proc (n)

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

</lang>

MACRO-10

<lang MACRO-10> 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 </lang>

Mathematica/Wolfram Language

<lang Mathematica>x=0; While[True,Print[BaseForm[x,8];x++]</lang>

MATLAB / Octave

<lang Matlab> n = 0;

   while (1)
       dec2base(n,8)
       n = n+1; 
   end; </lang>

Or use printf: <lang Matlab> n = 0;

   while (1)
       printf('%o\n',n);
       n = n+1; 
   end; </lang>

If a predefined sequence should be displayed, one can use <lang Matlab> seq = 1:100;

   dec2base(seq,8)</lang>

or <lang Matlab> printf('%o\n',seq);</lang>

Mercury

<lang>

- 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).

</lang>

min

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. <lang min>(

 (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</lang>

МК-61/52

Modula-2

<lang modula2>MODULE octal;

IMPORT InOut;

VAR num : CARDINAL;

BEGIN

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

END octal.</lang>

Nanoquery

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. <lang nanoquery>i = 0 while i < 2^64

   println format("%o", i)
   i += 1

end</lang>

NetRexx

<lang NetRexx>/* 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

</lang>

NewLISP

<lang NewLISP>; 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)</lang>

Sample output:

Nim

<lang nim>import strutils for i in 0 ..< int.high:

 echo toOct(i, 16)</lang>

Oberon-2

Works with: oo2c

<lang oberon2> 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. </lang>

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

<lang ocaml>let () =

 for i = 0 to max_int do
   Printf.printf "%o\n" i
 done</lang>

Output:

PARI/GP

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: <lang parigp>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++)</lang>

Automatic:

Works with: PARI/GP version 2.4.3 and above

<lang parigp>n=0;while(1,printf("%o\n",n);n++)</lang>

Pascal

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. <lang pascal>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. </lang>

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

Perl

Since task says "system register", I take it to mean "no larger than machine native integer limit": <lang perl>use POSIX; printf "%o\n", $_ for (0 .. POSIX::UINT_MAX);</lang> Otherwise: <lang perl>use bigint; my $i = 0; printf "%o\n", $i++ while 1</lang> 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. <lang perl>#!/usr/bin/perl

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

Phix

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

<lang php><?php for ($n = 0; is_int($n); $n++) {

 echo decoct($n), "\n";

} ?></lang>

Picat

Ways to convert to octal numbers:

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

 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).</lang>

Output:

0
1
2
3
4
5
6
7
10
11
...
17615737040105402212262317777777776
17615737040105402212262317777777777
17615737040105402212262320000000000
17615737040105402212262320000000001
17615737040105402212262320000000002
<Ctrl-C>

PicoLisp

<lang PicoLisp>(for (N 0 T (inc N))

  (prinl (oct N)) )</lang>

Pike

<lang Pike> int i=1; while(true)

   write("0%o\n", i++);

</lang>

Output:

01
02
...

PL/I

Version 1: <lang pli>/* 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;</lang> Version 2: <lang pli>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;</lang>

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/M

<lang plI>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</lang>

PowerShell

<lang PowerShell>[int64]$i = 0 While ( $True )

   {
   [Convert]::ToString( ++$i, 8 )
   }</lang>

Prolog

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.

<lang Prolog>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) ).</lang>

PureBasic

<lang PureBasic>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 </lang> Sample output:

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

Python

<lang Python>import sys for n in xrange(sys.maxint):

   print oct(n)</lang>

Quackery

<lang quackery>8 base put 0 [ dup echo cr 1+ again ]</lang>

Racket

lang racket

(for ([i (in-naturals)])

 (displayln (number->string i 8)))

</lang> (Racket has bignums, so this loop will never end.)

Raku

(formerly Perl 6) <lang perl6>say .base(8) for ^Inf;</lang>

Output:

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

REXX

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. <lang rexx>/*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. */</lang>

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

<lang ring> size = 30 for n = 1 to size

   see octal(n) + nl

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

</lang>

Ruby

From the documentation: "A Fixnum holds Integer values that can be represented in a native machine word (minus 1 bit). If any operation on a Fixnum exceeds this range, the value is automatically converted to a Bignum."

<lang ruby>n = 0 loop do

 puts "%o" % n
 n += 1

end

or

for n in 0..Float::INFINITY

 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</lang>

Run BASIC

<lang runbasic>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</lang>

Rust

<lang rust>fn main() {

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

}</lang>

Salmon

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 $2^{80}$ octal digits).

<lang Salmon>iterate (i; [0...+oo])

   printf("%o%\n", i);;</lang>

or

<lang Salmon>for (i; 0; true)

   printf("%o%\n", i);;</lang>

or

<lang Salmon>variable i := 0; while (true)

 {
   printf("%o%\n", i);
   ++i;
 };</lang>

S-BASIC

Although many (most?) BASICs have an OCT$ function, S-BASIC does not, so we have to supply our own <lang BASIC> 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 </lang>

Output:

Scala

<lang scala>Stream from 0 foreach (i => println(i.toOctalString))</lang>

Scheme

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

Scratch

Seed7

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

<lang seed7>$ 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;</lang>

Sidef

Simula

<lang simula> 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. </lang>

Smalltalk

Works with: Smalltalk/X

<lang smalltalk>0 to:Integer infinity do:[:n |

   n printOn:Stdout radix:8.
   Stdout cr.

]</lang>

Output:

Sparkling

<lang sparkling>for (var i = 0; true; i++) {

   printf("%o\n", i);

}</lang>

Standard ML

<lang sml>local

 fun count n = (print (Int.fmt StringCvt.OCT n ^ "\n"); count (n+1))

in

 val _ = count 0

end</lang>

Swift

<lang swift>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)

}</lang>

Output:

The first few lines. anyway:

Tcl

<lang tcl>package require Tcl 8.5; # arbitrary precision integers; we can count until we run out of memory! while 1 {

   puts [format "%llo" [incr counter]]

}</lang>

UNIX Shell

We use the bc calculator to increment our octal counter:

<lang sh>#!/bin/sh num=0 while true; do

 echo $num
 num=`echo "obase=8;ibase=8;$num+1"|bc`

done</lang>

Using printf

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

<lang sh>#!/bin/sh num=0 while test 0 -le $num; do

 printf '%o\n' $num
 num=`expr $num + 1`

done</lang>

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

<lang sh>num=0 while test 0 -le $num; do

 printf '%o\n' $num
 num=$((num + 1))

done</lang>

VBA

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

<lang VBA> 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 </lang>

VBScript

<lang vb> For i = 0 To 20 WScript.StdOut.WriteLine Oct(i) Next </lang>

Vim Script

<lang vim>let counter = 0 while counter >= 0

   echon printf("%o\n", counter)
   let counter += 1

endwhile</lang>

Vlang

<lang vlang>import math fn main() {

   for i := i8(0); ; i++ {
       println("${i:o}")
       if i == math.max_i8 {
           break
       }
   }

}</lang>

Output:

VTL-2

Stops at 65535, the largest integer supported by VTL-2. <lang VTL2>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</lang>

Output:

Whitespace

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

</lang>

It was generated from the following pseudo-Assembly.

<lang asm>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</lang>

Wren

Library: Wren-fmt

<lang ecmascript>import "/fmt" for Conv

var i = 0 while (true) {

   System.print(Conv.oct(i))
   i = i + 1

}</lang>

Output:

XPL0

XPL0 doesn't have built-in routines to handle octal; instead it uses hex. <lang XPL0>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; ]</lang>

Example output:

zig

<lang zig>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);
   }

}</lang>

Z80 Assembly

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.)

<lang z80> 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!</lang>

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

<lang zkl>foreach n in ([0..]){println("%.8B".fmt(n))}</lang>

Output:

ZX Spectrum Basic

<lang zxbasic>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</lang>