Count in octal

The task is to 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.

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

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:

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

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>

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.

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>

include <iomanip>

int main() {

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

}</lang>

Common Lisp

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

D

<lang d>import std.stdio;

void main() {

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

}</lang>

Delphi

<lang Delphi>program CountingInOctal;

{$APPTYPE CONSOLE}

uses SysUtils;

function DecToOct(aValue: Integer): string; var

 lRemainder: Integer;

begin

 Result := ;
 while aValue > 0 do
 begin
   lRemainder := aValue mod 8;
   Result := IntToStr(lRemainder) + Result;
   aValue := aValue div 8;
 end;

end;

var

 i: Integer;

begin

 for i := 1 to 20 do
   WriteLn(DecToOct(i));

end.</lang>

Euphoria

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

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

end while</lang>

Output:

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>

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

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 [:,@:(":"0) 8&#.^:_1

  (>:[disp)^:_ ]0</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:

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>

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>

Lua

<lang lua>for l=1,2147483647 do

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

end</lang>

Mathematica

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

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>

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

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>

Perl 6

PHP

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

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

} ?></lang>

PicoLisp

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

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

PL/I

<lang PL/I> /* 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>

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>

Retro

Integers in Retro are signed.

<lang Retro>octal 17777777777 [ putn cr ] iter</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>

Scala

<lang scala>0 until Int.MaxValue foreach(i=> println(i toOctalString))</lang>

Scheme

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

Seed7

<lang seed7>$ include "seed7_05.s7i";

const proc: main is func

 local
   var integer: i is 0;
 begin
   repeat
     writeln(str(i, 8));
     incr(i);
   until FALSE;
 end func;</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 = n + 1

end</lang>

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>

If the shell has $(( ... )) arithmetic, then we can increment a decimal counter and use printf(1) to print it in octal. Our loop stops when the counter overflows to negative.

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>