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bashexit-codelocal-variablescommand-substitution

Exit code of command substitution in bash local variable assignment

How can I check the exit code of a command substitution in bash if the assignment is to a local variable in a function?
Please see the following examples. The second one is where I want to check the exit code.
Does someone have a good work-around or correct solution for this?

$ function testing { test="$(return 1)"; echo $?; }; testing
1
$ function testing { local test="$(return 1)"; echo $?; }; testing
0
Solution

  • 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.