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Bash arithmetic issue

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# 8  
Old 07-22-2016
Quote:
Originally Posted by elbrand
and I only wanted to point out that this is an inconsistent behaviour. But, ok, we have two different operators here: (()) and the 'declare' one.
The entire reason (( )) exists is it behaves differently. (( )) is a modern C-style expression where you can mix assignments and comparisons, with a return code like you'd expect of such expressions. It's the modern way to do math and comparison in Bourne shell.

If this is inconsistent with the math syntax of plain Bourne circa 1977, well, good. There's a reason it's considered depreciated for modern shells.
# 9  
Old 07-25-2016
You are right. It's the same behaviour as in C (and I like C). So, thank you for the discussion. You changed my mind.
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