Why does “local” discard the return code of a command?

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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

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