Re: How come math/arithmetic cannot work by set -x

[Lawrence Velázquez]

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