Bash Retrieve Exit Code of Subshell Using Local – How to

exitexit-statussubshell

This question is close to others – Can I get the exit code from a sub shell launched with $(command)?

However there are no solutions I've found that allows me to get an exit code from a sub shell when using local and eval as in this example…

test() {
> local WHY="$(eval "echo 'test'"; exit 3)"; echo $?
> }
test
0

Best Answer

This is simple: Do not use a single command but split:

test() {
    local why
    why="$(eval "echo 'test'"; exit 3)"; echo $?
}
test
3

The problem was that local is a builtin command with an own exit code...If you avoid that command at the same time as the variable assignment, you get the exit code from the subshell.

Related Solutions

Shell Script – How to Exit Shell Script from a Subshell

You could kill the original shell (kill $$) before calling exit, and that'd probably work. But:

it seems rather ugly to me
it will break if you have a second subshell in there, i.e., use a subshell inside a subshell.

Instead, you could use one of the several ways to pass back a value in the Bash FAQ. Most of them aren't so great, unfortunately. You may just be stuck checking for errors after each function call (-e has a lot of problems). Either that, or switch to Perl.

How to Make `local` Capture the Exit Code in Bash

#!/bin/bash
thing() {
   local foo=$(asjkdh) ret="$?"
   echo "$ret"
}

This will echo 127, the correct error code for "command not found".

You can use local to define more than one variable. So I just also create the local variable RET to capture the exit code of the subshell before local succeeds and sets $? to zero.

Best Answer

Related Solutions

Shell Script – How to Exit Shell Script from a Subshell

How to Make `local` Capture the Exit Code in Bash

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