Re: Return from function depending on number of parameters

[Chris Elvidge]

On 03/07/2020 10:39 pm, Lawrence Velázquez wrote:
> > On Jul 3, 2020, at 2:00 PM, Chris Elvidge <celvidge001@gmail.com> wrote:
> >
> > However 'N=0; echo $((!$N))' gives an error at the bash prompt.
> > 'echo $[!$N]' echo's 1 as expected.
> >
> > My question - is $[...] actually obsolete?
>
> It might tell you something that $[...] is not even mentioned in
> the man page for bash 3.2.57, which is decidedly not the current
> version.
>
> > If so, what should I use at the bash prompt to get the same effect?
>
>
> I expect that the error you encountered was caused by !$ expanding
> to the last word of the previous command and making the contents
> of $((...)) an invalid arithmetic expression. This didn't affect
> your scripts because history expansion is not enabled in non-interactive
> shells by default.
>
> Try inserting a space.
>
>         $ N=0; printf %s\\n "$((! $N))"
>         1
>
> You can even drop the $.
>
>         $ N=0; printf %s\\n "$((! N))"
>         1
>
> vq

Thanks for the suggestion.

--
Chris Elvidge
England