Exit code of command substitution in bash local variable assignment [duplicate]

Answer 1

If you look at the man file for local (which is actually just the BASH builtins man page), it is treated as its own command, which gives an exit code of 0 upon successfully creating the local variable. So local is overwriting the last-executed error code.

Try this:

function testing { local test; test="$(return 1)"; echo $?; }; testing

EDIT: I went ahead and tried it for you, and it works.