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Re: shouldn't it the comma operator has the lowerest precedence in the shell arithmetic expression?

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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
> 
> 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/

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Re: shouldn't it the comma operator has the lowerest precedence in the shell arithmetic expression? Chet Ramey <chet.ramey@case.edu> - 2019-10-02 09:34 -0400

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