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Re: shouldn't it the comma operator has the lowerest precedence intheshe
From: |
hkadeveloper |
Subject: |
Re: shouldn't it the comma operator has the lowerest precedence intheshell arithmetic expression? |
Date: |
Wed, 2 Oct 2019 22:50:13 +0800 |
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/
- Re: shouldn't it the comma operator has the lowerest precedence intheshell arithmetic expression?,
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