Question

Consider foo.sh:

#!/bin/bash
function foo() {
   source another.sh
   echo "This shouldn't be executed. Return code: $?"
   return 0
}

foo
echo "Return code: $?"

Then another.sh:

echo "Inside another.sh"
return 1

Running ./foo.sh prints:

Inside another.sh
This shouldn't be executed. Return code: 1
Return code: 0

Is there an alternative method to include a source file into another such that the return command would return from the function enclosing include command rather than from the command itself?

Answer 1

One option is to propagate the return code in foo.sh:

source another.sh || return $?

Then:

$ ./foo.sh
Inside another.sh
Return code: 1

Alternatively, exit from the whole script in another.sh:

exit 1

Then:

$ ./foo.sh
Inside another.sh
$ echo $?
1