CODEWITHSUNDEEP
bashfunctionexitsubshell
LiMar
LiMar

Reputation: 2972

How to effectively abort the execution of a Bash script from a function

I was using the "exit 1" statement in my Bash functions to terminate the whole script and it worked fine:

function func()
{
   echo "Goodbye"
   exit 1
}
echo "Function call will abort"
func
echo "This will never be printed"

But then I realized that it doesn't do the work when called like:

res=$(func)

I understand that I created a subshell and "exit 1" aborts that subshell and not the primary one; is there a way to write a function which aborts the whole execution, no matter how it is called?

I just need to get the real return value (echoed by the function).

Upvotes: 114

Views: 73123

Answers (8)

Reiner
Reiner

Reputation: 1

I think the easiest and cleanest is

kill –TERM $$

Upvotes: -2

LinuxBlack
LinuxBlack

Reputation: 1

This works and is "clean" without a lot of extra hoops, make sure the top level function is defined with parenthesis instead of curly braces. This provides 1) immediate exit as well as 2) the exit status is available:

TestExitAllFunctions ()
(
    function FatalError ()
    {
        echo "$1" 1>&2;
        exit 1
    };
    echo "Function call will abort";
    FatalError "Oops, fatal error found, sorry";
    echo "This will never be printed"
)

TestExitAllFunctions
Function call will abort
Oops, fatal error found, sorry

echo $?
1

Upvotes: -1

John
John

Reputation: 21

If you just need top be able to bomb out of your scripts from within a function, you can do:

function die () {
    set -e
    /bin/false
}

then elsewhere, within your functions, instead of using "exit", use "die".

Upvotes: 2

AmazingAlex
AmazingAlex

Reputation: 29

I guess better is 

#!/bin/bash
set -e
trap "exit 1" ERR

myfunc() {
     set -x # OPTIONAL TO SHOW ERROR
     echo "Exit with failure"
     set +x # OPTIONAL
     exit 1
}
echo "BEFORE..."
myvar="$(myfunc)"
echo "AFTER..But not shown"

Upvotes: 2

brice
brice

Reputation: 25039

You can use set -e which exits if a command exits with a non-zero status:

set -e 
func
set +e

Or grab the return value:

(func) || exit $?

Upvotes: 57

FatalError
FatalError

Reputation: 54611

What you could do, is register the top level shell for the TERM signal to exit, and then send a TERM to the top level shell:

#!/bin/bash
trap "exit 1" TERM
export TOP_PID=$$

function func()
{
   echo "Goodbye"
   kill -s TERM $TOP_PID
}

echo "Function call will abort"
echo $(func)
echo "This will never be printed"

So, your function sends a TERM signal back to the top level shell, which is caught and handled using the provided command, in this case, "exit 1".

Upvotes: 129

Luca
Luca

Reputation: 4273

But is there a way to write a function which aborts the whole execution, no matter how it is called?

No.

I just need to get the real return value (echoed by the function).

You can 

res=$(func)
echo $?

Upvotes: 0

Daenyth
Daenyth

Reputation: 37461

A child process can't force the parent process to close implicitly. You need to use some kind of signaling mechanism. Options might include a special return value, or perhaps sending some signal with kill, something like

function child() {
    local parent_pid="$1"
    local other="$2"
    ...
    if [[ $failed ]]; then
        kill -QUIT "$parent_pid"
    fi
}

Upvotes: 3

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