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Re: How come math/arithmetic cannot work by set -x
From: |
Lawrence Velázquez |
Subject: |
Re: How come math/arithmetic cannot work by set -x |
Date: |
Fri, 12 Aug 2022 20:23:53 -0400 |
User-agent: |
Cyrus-JMAP/3.7.0-alpha0-841-g7899e99a45-fm-20220811.002-g7899e99a |
On Fri, Aug 12, 2022, at 7:40 PM, Dennis Williamson wrote:
> It works for me. What are you expecting?
>
> It would help if you show what you're doing, the result you're getting and
> what you expect instead.
I'm guessing that instead of, for example
% bash -xc 'a="(x=1)" b="2*3"; ((a+b))'
+ a='(x=1)'
+ b='2*3'
+ (( a+b ))
they want something like
% bash -xc 'a="(x=1)" b="2*3"; ((a+b))'
+ a='(x=1)'
+ b='2*3'
+ (( a+b ))
+ (( (x=1)+2*3 ))
+ (( 1+6 ))
+ (( 7 ))
or whatever. Who knows.
--
vq