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	shouldn't it the comma operator has the lowerest precedence in the shell arithmetic expression?
Date	2019-10-02 08:35 +0800
Message-ID	<mailman.707.1569976552.2651.bug-bash@gnu.org> (permalink)
References	<CAD0rTC4c80xV7VtUehFm1RYycGFFdO2Mun7C8qw_PP9QgZHVCg@mail.gmail.com>

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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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shouldn't it the comma operator has the lowerest precedence in the shell arithmetic expression? hk <hkadeveloper@gmail.com> - 2019-10-02 08:35 +0800

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