# Count in octal

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

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.

## 0815

`}: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

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

`*        Octal                     04/07/2016OCTAL    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=0LOOPI    LR     R2,R6              x=i         LA     R9,10              j=10         LA     R4,PG+23           @pgLOOP     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               @[email protected]         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 iELOOPI   L      R13,4(0,R13)       epilog          LM     R14,R12,12(R13)    "         XR     R15,R15            "         BR     R14                exit         LTORG  PG       DC     CL80' '            bufferDW       DS     0D,PL8             15numZN12     DS     CL12EM12     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
```

`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

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

## ALGOL 68

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

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_MAXINTEGERend.`
Output:
```0
1
2
3
4
5
6
7
10
11
12
...
```

## AutoHotkey

`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

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

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

Works with: QBasic
`DIM n AS LONGFOR 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 + 1WENDENDFUNCTION 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`

### Applesoft BASIC

`10 N\$ = "0" 100 O\$ = N\$110 PRINT O\$120 N\$ = ""130 C = 1140 FOR I = LEN(O\$) TO 1 STEP -1150     N = VAL(MID\$(O\$, I, 1)) + C160     C = N >= 8170     N\$ = STR\$(N - C * 8) + N\$180 NEXT I190 IF C THEN N\$ = "1" + N\$200 GOTO 100`

### 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.

` 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 130110 LET L=L-1120 GOTO 50130 LET N\$="1"+N\$140 GOTO 20`

## Batch File

` @echo off:: {CTRL + C} to exit the batch file:: Send incrementing decimal values to the :to_Oct functionset loop=0:loop1call:to_Oct %loop%set /a loop+=1goto loop1:: Convert the decimal values parsed [%1] to octal and output them on a new line:to_Octset todivide=%1set "fulloct=" :loop2set tomod=%todivide%set /a appendmod=%tomod% %% 8set fulloct=%appendmod%%fulloct%if %todivide% lss 8 (  echo %fulloct%  exit /b)set /a todivide/=8goto loop2 `
Output:
```0
1
2
3
4
5
6
7
10
...
```

## BBC BASIC

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

`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.

## 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.

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

## Bracmat

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***

`+[            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

`#include <stdio.h> int main(){        unsigned int i = 0;        do { printf("%o\n", i++); } while(i);        return 0;}`

## C#

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

## 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.

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

## Clojure

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

## COBOL

Translation of: Delphi
Works with: GNU Cobol version 2.0
`       >>SOURCE FREEIDENTIFICATION 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

` n = 0 while true  console.log n.toString(8)  n += 1 `

## Common Lisp

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

## Component Pascal

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	ENDEND 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
```

## Crystal

`# 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

`void main() {    import std.stdio;     ubyte i;    do writefln("%o", i++);    while(i);}`

## Dc

A simple infinite loop and octal output will do.

`8o0[p1+lpx]dspx`

## DCL

`\$ 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

`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.`

## Elixir

`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) endf.(f, 0)`

## Emacs Lisp

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

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

## Erlang

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

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

Output:

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

## F#

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

## Factor

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

## Forth

Using INTS from Integer sequence#Forth

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

## Fortran

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 doend program`

## FreeBASIC

`' FB 1.05.0 Win64 Dim ub As UByte = 0 ' only has a range of 0 to 255Do   Print Oct(ub, 3)   ub += 1Loop Until ub = 0 ' wraps around to 0 when reaches 256PrintPrint "Press any key to quit"Sleep`

## Futhark

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

` include "ConsoleWindowdefstr word dim as short i for i = &o000000 to &o000031      // 0 to 25 in decimal   print oct\$(i); " in octal ="; inext `

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

`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

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
...```

`import Numeric main = mapM_ (putStrLn . flip showOct "") [1..]`

## Icon and Unicon

`link convert   # To get exbase10 method procedure main()    limit := 8r37777777777    every write(exbase10(seq(0)\limit, 8))end`

## J

Solution:

`   disp=.([smoutput) ' '(-.~":)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)^:_]0x012345671011...`

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

`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

`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}`

## Julia

` 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
```

## Kotlin

`//  version 1.1 //  counts up to 177 octal i.e. 127 decimalfun 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

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

## Lang5

`'%4o '__number_format set0 do dup 1 compress . "\n" . 1 + loop`

## LFE

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

## Liberty BASIC

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 = 0while 1    call CountOctal 0, i, i > 0    i = i + 1wend 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 iend sub `

## 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.

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

## LOLCODE

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 octalizerIF U SAY SO IM IN YR printer UPPIN YR num    VISIBLE I IZ octal YR num MKAYIM OUTTA YR printer KTHXBYE`

## Lua

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

## Maple

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

## Mathematica

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

## MATLAB / Octave

`    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

` :- 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). `

## МК-61/52

`ИП0	П1	1	0	/	[x]	П1	Вx	{x}	10	*	7	-	x=0	21	ИП1	x#0	28	БП02	ИП0	1	+	П0	С/П	БП	00	ИП0	lg[x]	1	+	10^x	П0	С/П	БП	00`

## Modula-2

`MODULE octal; IMPORT  InOut; VAR     num             : CARDINAL; BEGIN  num := 0;  REPEAT    InOut.WriteOct (num, 12);           InOut.WriteLn;    INC (num)  UNTIL num = 0END octal.`

## 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 = 8k_ = BigIntegerk_ = BigInteger.ZERO loop forever  say k_.toString(int radix)  k_ = k_.add(BigInteger.ONE)  end `

## 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)`

Sample output:

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

## Nim

`import strutilsfor i in 0 .. <int64.high:  echo toOct(i, 16)`

## Oberon-2

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  ENDEND 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

`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

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

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;//8^9var  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```

## Perl

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`

## 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. :-)

## Phix

`integer i = 0constant ESC = #1Bwhile not find(get_key(),{ESC,'q','Q'}) do    printf(1,"%o\n",i)    i += 1end while `

## PHP

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

## PicoLisp

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

## PL/I

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
```

## PowerShell

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

## 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.qIf 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

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

## Racket

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

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

## Retro

Integers in Retro are signed.

`octal17777777777 [ putn cr ] iter`

## REXX

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

`/*REXX program counts in octal until the number exceeds #pgm 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 all big numbers. */numIn=right('number in', 20)           /*used for indentation of output.*/w=length(sourceline())                 /*used for formatting width of #s*/   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 #protons>pgm statements*/  end   /*#*/                                       /*stick a fork in it, we're 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
```

## Ring

` size = 30for n = 1 to size    see octal(n) + nlnext 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

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."

`n = 0loop do  puts "%o" % n  n += 1end # orfor n in 0..Float::INFINITY  puts n.to_s(8)end # or0.upto(1/0.0) do |n|  printf "%o\n", nend # version 2.1 later0.step do |n|  puts format("%o", n)end`

## Run BASIC

`input "Begin number:";binput "  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 fornext iend function`

## Rust

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

## 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 ${\displaystyle 2^{80}}$ 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;  };`

## Scala

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

## Scheme

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

## Seed7

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

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

## Sparkling

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

## Standard ML

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

## 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)}`
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

`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

We use the bc calculator to increment our octal counter:

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

### Using printf

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

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

## VBA

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 Integeri = 0On Error GoTo OctEndDo    Debug.Print Oct(i)    i = i + 1LoopOctEnd:Debug.Print "Integer overflow - count terminated"End Sub `

## VBScript

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

## Vim Script

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

## 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.

`           		 	  	  	 	   	 	 	     		          	     	   	 		 	   	   	 	      	  		  			 	      	   		 			 	  	     		   	`

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 23:    pop    ret`

## XPL0

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 octalint 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
```

## zkl

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

## ZX Spectrum Basic

`10 PRINT "DEC.  OCT."20 FOR i=0 TO 2030 LET o\$="": LET n=i40 LET o\$=STR\$ FN m(n,8)+o\$50 LET n=INT (n/8)60 IF n>0 THEN GO TO 4070 PRINT i;TAB 3;" = ";o\$80 NEXT i90 STOP 100 DEF FN m(a,b)=a-INT (a/b)*b`