bashshelllocal

Why does "local" discard the return code of a command?

This Bash snippet works as expected:

$ fun1() { x=$(false); echo "exit code: $?"; }
$ fun1
exit code: 1

But this one, using local, does not as I would have expected:

$ fun2() { local x=$(false); echo "exit code: $?"; }
$ fun2
exit code: 0

Can anyone explain why does local discards the return code of the command?

Solution

  • The reason the code with local returns 0 is because $? "Expands to the exit status of the most recently executed foreground pipeline." Thus $? is returning the success of local

    You can fix this behavior by separating the declaration of x from the initialization of x like so:

    $ fun() { local x; x=$(false); echo "exit code: $?"; }; fun
exit code: 1