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Re: shouldn't it the comma operator has the lowerest precedence in the s
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From: |
Chet Ramey |
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Subject: |
Re: shouldn't it the comma operator has the lowerest precedence in the shell arithmetic expression? |
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Date: |
Wed, 2 Oct 2019 09:34:08 -0400 |
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User-agent: |
Mozilla/5.0 (Macintosh; Intel Mac OS X 10.14; rv:60.0) Gecko/20100101 Thunderbird/60.9.0 |
On 10/1/19 8:35 PM, hk wrote:
> Configuration Information :
> Bash Version: 5.0
> Patch Level: 0
> Release Status: release
>
> Description:
> 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 chet@case.edu http://tiswww.cwru.edu/~chet/