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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: Wed, 2 Oct 2019 08:35:30 +0800
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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?