$# positional parameters of what

Asked Jan 09 '20 at 06:00

Active Jan 09 '20 at 06:07

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I have the following bash script in my code.

#!/bin/sh
set -eu
[ $# -eq 1 ] || exit 1
[ "$1" = "wlanX" ] || exit 2
[ "$PATH" = "/path1:/path2" ] || exit 3
exit 0

I read that $# counts the number of positional parameters. But my question is positional parameters of what ? I am tyring to figure out what the above code means?

asked Jan 09 '20 at 06:00

liv2hak

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IMHO it is parameter passed to a script which `$#` takes care of. – RavinderSingh13 Jan 09 '20 at 06:07
You could check this link once https://stackoverflow.com/questions/36007240/what-is-the-use-of-in-bash is this helps you too. – RavinderSingh13 Jan 09 '20 at 06:08

$# positional parameters of what

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