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

[Dennis Williamson]

On Fri, Aug 12, 2022 at 6:51 PM Budi <budikusasi@gmail.com> wrote:

> It doesn't work means no use on set -x, no value is shown
>
> On 8/13/22, Dennis Williamson <dennistwilliamson@gmail.com> wrote:
> > On Fri, Aug 12, 2022, 6:28 PM Budi <budikusasi@gmail.com> wrote:
> >
> >> How come math/arithmetic ((i=k+l)) cannot make use of set -x
> >>
> >> Please help..
> >> (so annoying).
> >>
> >
> >
> > 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.
> >
> > "It doesn't work" conveys no information whatsoever.
> >
> >>
> >
>

Hmmm... interesting.

$ set -x; unset a; b=2; c=7; ((a = $b + $c)); echo "$a $b $c"; set +x

+ unset a
+ b=2
+ c=7
+ (( a = 2 + 7 ))
+ echo '9 2 7'
9 2 7
+ set +x

shows the values in the arithmetic expression, but

set -x; unset a; b=2; c=7; ((a = b + c)); echo "$a $b $c"; set +x

+ unset a
+ b=2
+ c=7
+ (( a = b + c ))
+ echo '9 2 7'
9 2 7
+ set +x

without the dollar signs doesn't.

Also, note that this message includes a thorough example of doing, result,
expecting.

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