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Re: Return from function depending on number of parameters

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  Re: Return from function depending on number of parameters Lawrence Velázquez <vq@larryv.me> - 2020-07-03 17:39 -0400

#16530 — Re: Return from function depending on number of parameters

> 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

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