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 is obsolete and that $((...))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".)