Returning value from called function in a shell script

Question

I want to return the value from a function called in a shell script. Perhaps I am missing the syntax. I tried using the global variables. But that is also not working. The c

Answer 1

A Bash function can't return a string directly like you want it to. You can do three things:

1. Echo a string
2. Return an exit status, which is a number, not a string
3. Share a variable

This is also true for some other shells.

Here's how to do each of those options:

1. Echo strings

lockdir="somedir"
testlock(){
    retval=""
    if mkdir "$lockdir"
    then # Directory did not exist, but it was created successfully
         echo >&2 "successfully acquired lock: $lockdir"
         retval="true"
    else
         echo >&2 "cannot acquire lock, giving up on $lockdir"
         retval="false"
    fi
    echo "$retval"
}

retval=$( testlock )
if [ "$retval" == "true" ]
then
     echo "directory not created"
else
     echo "directory already created"
fi

2. Return exit status

lockdir="somedir"
testlock(){
    if mkdir "$lockdir"
    then # Directory did not exist, but was created successfully
         echo >&2 "successfully acquired lock: $lockdir"
         retval=0
    else
         echo >&2 "cannot acquire lock, giving up on $lockdir"
         retval=1
    fi
    return "$retval"
}

testlock
retval=$?
if [ "$retval" == 0 ]
then
     echo "directory not created"
else
     echo "directory already created"
fi

3. Share variable

lockdir="somedir"
retval=-1
testlock(){
    if mkdir "$lockdir"
    then # Directory did not exist, but it was created successfully
         echo >&2 "successfully acquired lock: $lockdir"
         retval=0
    else
         echo >&2 "cannot acquire lock, giving up on $lockdir"
         retval=1
    fi
}

testlock
if [ "$retval" == 0 ]
then
     echo "directory not created"
else
     echo "directory already created"
fi