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| From | Chet Ramey <chet.ramey@case.edu> |
| Newsgroups | gnu.bash.bug |
| Subject | Re: shouldn't it the comma operator has the lowerest precedence in the shell arithmetic expression? |
| Date | Wed, 2 Oct 2019 09:34:08 -0400 |
| Lines | 27 |
| Approved | bug-bash@gnu.org |
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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/
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Re: shouldn't it the comma operator has the lowerest precedence in the shell arithmetic expression? Chet Ramey <chet.ramey@case.edu> - 2019-10-02 09:34 -0400
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