Exit a bash script from a function whose output is captured

Question

I want to completely terminate/exit a bash shell script upon error, but using a function error that lets me display a debug output before termination. Now I have the problem that the exit 1 statement inside the error function will not terminate the shell script if the function's output is captured via backticks or $().

Here is my sample script:

    #!/bin/bash

    function error ()
    {
            echo "An error has occured: $1"
            exit 1
    }
    function do_sth ()
    {
            if [ $1 -eq 0 ]; then
                    error "First param must be greater than 0!"
            else
                    echo "OK!"
            fi
    }

    RESULT=`do_sth 0`

    echo "This line should never be printed"

How can I immediately terminate the script in the error() function?

Olaf Dietsche · Accepted Answer · 2013-01-17 12:27:26Z

5

The problem with command substitution is, that a subshell is started to execute do_sth. exit 1 then terminates this subshell and not the main bash.

You can work around this by appending || exit $?, which exits with the exit code from the command substitution

RESULT=`do_sth 0` || exit $?

If you want to show the error message, redirect it to stderr

echo "An error has occured: $1" >&2

edited Jan 17, 2013 at 12:27

answered Jan 17, 2013 at 12:15

Olaf Dietsche

74.4k9 gold badges113 silver badges214 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Dynalon Over a year ago

Problem with that solution is, that the actual error message generated by the error function will not be printed

J Homard · Accepted Answer · 2013-01-17 14:38:00Z

1

RESULT=`do_sth 0` || exit $?

then echo "An error has occured: $1" >&2

answered Jan 17, 2013 at 14:38

J Homard

3122 gold badges3 silver badges9 bronze badges

Comments

William Pursell · Accepted Answer · 2013-01-17 13:41:39Z

There's no way to avoid the fact that the subshell cannot make the parent terminate directly: the parent will have to evaluate the value returned from the subshell. A common technique is to trap exit and print an error message in the trap function. Since you are generating the error message in a subshell, you cannot simply assign the message to a variable as otherwise might be done, but you can use the filesystem. Please note that this idea is really silly, and it is much cleaner to simply write error messages to stderr. That's what it is for, and that's why it is inherited by children. Something like:

#!/bin/sh

trap final 0 2 15

# Create a temporary file to store error messages.  This is a terrible
# idea: it would be much better to simply write error messages to stderr,
# but this code is attempting to demonstrate the technique of having the
# parent print the message.  Perhaps it would do better to serve as an example
# of why reporting your children's mistakes is a bad idea.  The children
# should be responsible for reporting their own errors.  Doing so is easy,
# since they inherit file descriptor 2 from their parent.
errmsg=$( mktemp xxxxx )
final() {
    test "$?" = 0 || cat $errmsg
    rm -f $errmsg
} >&2

# Must emphasize one more time that this is a silly idea.  The error 
# function ought to be writing to stderr: eg echo "error: $*" >&2
error() { echo "error: $*" > $errmsg; exit 1; }
do_sth() { 
    if test "$1" -eq 0; then
        error "First param must be greater than 0!"
    else
        echo "OK!"
    fi
}


result=$( do_sth 0 ) || exit 1
echo not printed

Community · Accepted Answer · 2017-05-23 10:27:19Z

0

It seems the (clever) answer is on another question's top answer by @FatalError, here on so

edited May 23, 2017 at 10:27

CommunityBot

11 silver badge

answered Jan 17, 2013 at 14:05

Olivier Dulac

3,84019 silver badges35 bronze badges

Comments

anishsane · Accepted Answer · 2013-01-17 14:56:22Z

0

This should work...

#!/bin/bash

trap "exit 1" 50                                       #exit process after receiving signal 50.

function myerror ()
{
    echo "An error has occured: $1" >&2
}
function do_sth ()
{
    if [ $1 -eq 0 ]; then
        myerror "First param must be greater than 0!"
        kill -50 $(ps --pid $$ -opid=)                 #uncommon signal 50 is used.
    else
        echo "OK!"
    fi
}

RESULT=`do_sth 1`
echo $RESULT

RESULT=`do_sth 0`
echo $RESULT

echo "This line should never be printed"

answered Jan 17, 2013 at 14:56

anishsane

21.1k5 gold badges46 silver badges77 bronze badges

1 Comment

anishsane Over a year ago

You can have multiple such signals like 50 used here & thus different exit statuses...

Collectives™ on Stack Overflow

Exit a bash script from a function whose output is captured

5 Answers 5

1 Comment

Comments

Comments

Comments

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

5 Answers 5

1 Comment

Comments

Comments

Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related