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

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   Yes. I understand what you are saying. I mean isn’t it a little
   inconsistent about the comment, the macro name(EXP_HIGHEST) and the
   macro value(expcomma)?

   Thanks for you reply!

   ------------------ Original ------------------
   From: "Chet Ramey" <chet.ramey@case.edu>; <"Chet Ramey"
   <chet.ramey@case.edu>;>
   Date: Wed,Oct 2,2019 10:28 PM
   To: hkadeveloper <hkadeveloper@gmail.com>
   Subject: Re: shouldn't it the comma operator has the lowerest
   precedence inthe shell arithmetic expression?
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 intheshell arithmetic expression? "hkadeveloper"<hkadeveloper@gmail.com> - 2019-10-02 22:50 +0800

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