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.ra...@case.edu>; <"Chet Ramey"
<chet.ra...@case.edu>;>
Date: Wed,Oct 2,2019 10:28 PM
To: hkadeveloper <hkadevelo...@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 c...@case.edu http://tiswww.cwru.edu/~chet/
