Re: Exit application with two function calls

[Bernd Eggink]

abi@e-arroyo.net schrieb:
> Hi,
>
> I am working on a script and ran into an unusual program. Consider the
> following script which I called how.sh.
>
> =====
>
> #!/bin/bash
>
> error () {
>
>    echo -e "\n\terror: ${*}\n"
>    exit;
>
>    # kill $$
>
> }
>
> check_file () {
>
>    input="$*"
>
>    if [ -e $input ]; then
>       echo "$input"
>    else
>       error "invalid directory: $input"
>    fi
>
> }
>
>
> chkFile="`check_file $1`"
> echo $chkFile
> echo "don't print"
>
> =====
>
> When I run the command above I get this output:
>
> # ./how.sh /asdf
> error: invalid directory: /asdf
> don't print

The reason is that `check_file` is executed in a subshell, so 'exit' 
just leaves the subshell, not the script itself.

The command substitution is unnecessary anyway, as the result (if any) 
will simply be $1 itself. Thus, if you replace the main part by	

    check_file $1
    echo $1

it should work as expected.

Regards,
Bernd

--
Bernd Eggink
monoped@sudrala.de
http://sudrala.de