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

BashShellLocal

Bash Problem Overview

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 sweep the return code of the command?

Bash Solutions

Solution 1 - Bash

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

Solution 2 - Bash

The return code of the local command obscures the return code of false

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestiontoklandView Question on Stackoverflow
Solution 1 - BashSiegeXView Answer on Stackoverflow
Solution 2 - BashIgnacio Vazquez-AbramsView Answer on Stackoverflow