Re: Return from function depending on number of parameters

[Eli Schwartz]

On 7/3/20 2:00 PM, Chris Elvidge wrote:
> I've used 'return $((!$#))' and 'return $[!$#]' to return an error if no
> parameters given to function.
> 
> Tested in a bash script 'exit $((!$#)) / $[!$#]' - both work.
> 
> 'echo  $((!$#)) / $[!$#]' - both echo 1 when no params, 0 when any
> number of params.
> 
> I'm told ( https://wiki.bash-hackers.org/scripting/obsolete ) that
> $[...] is obsolete and that $((...)) should be used instead. OK so far.
> 
> However 'N=0; echo $((!$N))' gives an error at the bash prompt. 'echo
> $[!$N]' echo's 1 as expected.

"gives an error" is a useless bug report. It works for me.

$ N=0; echo $((!$N))
1

My initial reaction to reading this thread is head scratching!

As the other reply mentioned, there's actually a good explanation for
why we get different results -- I disabled an annoying feature.

$ set -o histexpand

Now here's a useful bug report. "When I run this, I get the following
incorrect results or error message":

$ N=0; echo $((!$N))
N=0; echo $((histexpandN))
0
$ N=0; echo $((!$N))
N=0; echo $(()N))
-bash: syntax error near unexpected token `)'
$ N=0
$ echo $((!$N))
echo $((N=0N))
-bash: N=0N: value too great for base (error token is "0N")

...

From there, people can give useful advice for solving the problem. (My
preferred advice is "disable histexpand".)

-- 
Eli Schwartz
Arch Linux Bug Wrangler and Trusted User

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