exit a bash script from a command substitution

Question

Is it possible to exit a script from a function that is called using command substitution?

e.g

#/bin/bash
function do_and_check()
{
    ls $1 || exit
}

p=$(do_and_check /etc/passwd)
q=$(do_and_check /xxx)

the result of running this is

bash -x yyy                                                                                                             
++ do_and_check /etc/passwd                                                                                                                  
++ ls /etc/passwd                                                                                                                             
+ p=/etc/passwd
++ do_and_check /xxx
++ ls /xxx
ls: cannot access /xxx: No such file or directory
++ exit
+ q=
+ echo /etc/passwd
/etc/passwd

I would have liked the exit to take me out of the script, not out of the command substitution. Is that possible?

Brian Agnew · Accepted Answer · 2013-01-03 09:42:38Z

3

This:

p=$(do_and_check /etc/passwd)

is executing a subshell, and that's what you're exiting from. I would set the exit value in your function (to a non-zero value), and then check the return value $? after the command substitution.

answered Jan 3, 2013 at 9:42

Brian Agnew

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2 Comments

glenn jackman Over a year ago

+1. Or, use set -e to have the script exit on any non-zero return status.

dbbd Over a year ago

adding a $? check after each call to the function is what I want to avoid in the first place. set -e is too encompassing. I can call directly e.g. do_and_check xxx and have it set a known variable ala perl's $_; or pehraps do_and_check could kill its parent?

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