Count in octal: Difference between revisions

From Rosetta Code
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({{header|Lua}})
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print(string.format("%o",l))
print(string.format("%o",l))
end</lang>
end</lang>

=={{header|Modula-2}}==

<lang Modula-2>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>


=={{header|Perl 6}}==
=={{header|Perl 6}}==

Revision as of 23:15, 7 June 2011

Count in octal is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

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 that can be held within the system registers is reached (for a 32 bit system using unsigned registers, this value is 37777777777 octal).

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 = 1; l < 2147483647; l++) {
   printf("%o\n", l);
 }

}</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>

  1. include <iomanip>

using namespace std;

int main() {

 unsigned i = 0;
 do
 {
   cout << setbase(8) << i << endl;
   ++i;
 } while(i != 0);

}</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>

Java

<lang java>public class Count{

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

}</lang>

Lua

<lang lua>for l=1,2147483647 do

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

end</lang>

Modula-2

<lang Modula-2>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>

Perl 6

<lang perl6>say .fmt: '%o' for 0 .. *;</lang>

PicoLisp

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

  (prinl (oct N)) )</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>