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Re: shouldn't it the comma operator has the lowerest precedence in the s


From: Chet Ramey
Subject: Re: shouldn't it the comma operator has the lowerest precedence in the shell arithmetic expression?
Date: Wed, 2 Oct 2019 09:34:08 -0400
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    address@hidden    http://tiswww.cwru.edu/~chet/



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