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| From: | hk |
| 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 |
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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