Exit code of variable assignment to command substitution in Bash

Question

I am confused about what error code the command will return when executing a variable assignment plainly and with command substitution:

a=$(false); echo $?
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Answer 1

Note that this isn't the case when combined with local, as in local variable="$(command)". That form will exit successfully even if command failed.

Take this Bash script for example:

#!/bin/bash

function funWithLocalAndAssignmentTogether() {
    local output="$(echo "Doing some stuff.";exit 1)"
    local exitCode=$?
    echo "output: $output"
    echo "exitCode: $exitCode"
}

function funWithLocalAndAssignmentSeparate() {
    local output
    output="$(echo "Doing some stuff.";exit 1)"
    local exitCode=$?
    echo "output: $output"
    echo "exitCode: $exitCode"
}

funWithLocalAndAssignmentTogether
funWithLocalAndAssignmentSeparate

Here is the output of this:

nick.parry@nparry-laptop1:~$ ./tmp.sh 
output: Doing some stuff.
exitCode: 0
output: Doing some stuff.
exitCode: 1

This is because local is actually a builtin command, and a command like local variable="$(command)" calls local after substituting the output of command. So you get the exit status from local.