Why does my "exit" command not exit my Bash script?

Question

I have this Bash script:

#!/bin/bash
set -x
function doSomething() {
    callee
    echo "It should not go to here!"
}

function callee() {
    ( echo "before" ) && (echo "This is callee" && exit 1 )                                                                                   
    echo "why I can see this?"
}


doSomething

and this is the result:

+ set -x
+ doSomething
+ callee
+ echo before
before
+ echo 'This is callee'
This is callee
+ exit 1
+ echo 'why I can see this?'
why I can see this?
+ echo 'It should not go to here!'
It should not go to here!

I see the command exit, but it doesn't exit the script – why doesn't exit work?

chepner · Accepted Answer · 2012-09-04 14:52:28Z

6

You are calling exit from inside a subshell, so that's the shell that is exiting. Try this instead:

function callee() {
    ( echo "before" ) && { echo "This is callee" && exit 1; }                                                                                   
    echo "why I can see this?"
}

This, however, will exit from whatever shell called callee. You may want to use return instead of exit to return from the function.

answered Sep 4, 2012 at 14:52

chepner

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Comments

zigdon · Accepted Answer · 2012-09-04 14:52:48Z

3

When you run a command in () you're spawning a subshell. So when you call exit within that subshell, you're just exiting it, and not your top level script.

answered Sep 4, 2012 at 14:52

zigdon

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Comments

Kobor42 · Accepted Answer · 2012-09-04 14:53:03Z

2

Because round paredenthesis creates a new nested shell, which is exited with exit.

answered Sep 4, 2012 at 14:53

Kobor42

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