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

From	hk <hkadeveloper@gmail.com>
Newsgroups	gnu.bash.bug
Subject	Re: shouldn't it the comma operator has the lowerest precedence in the shell arithmetic expression?
Date	2019-10-02 08:59 +0800
Message-ID	<mailman.710.1569978016.2651.bug-bash@gnu.org> (permalink)
References	<CAD0rTC4c80xV7VtUehFm1RYycGFFdO2Mun7C8qw_PP9QgZHVCg@mail.gmail.com> <CAD0rTC5ivYg8xH7tF=TTa=31Hj1q-CNP+0xR6VvP3BkaZTb_Eg@mail.gmail.com>

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sorry for the typo in the title, it should be *lowest*.

On Wed, Oct 2, 2019 at 8:35 AM hk <hkadeveloper@gmail.com> 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?
>

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Re: shouldn't it the comma operator has the lowerest precedence in the shell arithmetic expression? hk <hkadeveloper@gmail.com> - 2019-10-02 08:59 +0800

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