Question

I’ve got following bash script to do something for each parameter of the script

#! /bin/sh

while (($#)); do
 echo $1
 shift
done

But somehow, if I start it with the command sudo ./test.sh foo1 foo2 it wont work. And the real strange thing is, that if I enter sudo bash test.sh foo1 foo2 it works. Does anybody know what causes this strange behaviour?

Answer 1

You have specified /bin/sh as your interpreter, which may not be bash. Even if it is bash, bash runs in POSIX mode when called as /bin/sh.

The (( )) command is a bash-specific feature. The following will work in any POSIX compliant shell:

while [ $# -gt 0 ]; do
   echo $1
   shift
done