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Re: shouldn't it the comma operator has the lowerest precedence in the s
Re: shouldn't it the comma operator has the lowerest precedence in the shell arithmetic expression?
Wed, 2 Oct 2019 09:34:08 -0400
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On 10/1/19 8:35 PM, hk wrote:
> Configuration Information :
> Bash Version: 5.0
> Patch Level: 0
> Release Status: release
> the code snippet from expr.c starting from line 141:
>> /* This should be the function corresponding to the operator with the
>> highest precedence. */
>> #define EXP_HIGHEST expcomma
> Am I understanding it wrong or is it a typo?
The bash arithmetic parser does things in reverse order, in a way. So
the comma operator is the first thing you call, and it calls functions
that implement the other operators in ascending priority order. You
didn't misunderstand it.
``The lyf so short, the craft so long to lerne.'' - Chaucer
``Ars longa, vita brevis'' - Hippocrates
Chet Ramey, UTech, CWRU address@hidden http://tiswww.cwru.edu/~chet/