Count in octal: Difference between revisions
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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).
=={{header|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>
=={{header|UNIX Shell}}==
|
Revision as of 00:03, 6 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).
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>
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>