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). |
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). |
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=={{header|UNIX Shell}}== |
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We use the bc calculator to increment our octal counter: |
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<lang sh>#!/bin/sh |
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num=0 |
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while true; do |
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echo $num |
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num=`echo "obase=8;ibase=8;$num+1"|bc` |
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done</lang> |
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[[Category:Basic language learning]] |
[[Category:Basic language learning]] |
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Revision as of 23:58, 5 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).
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>