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 $?
<         


        

Answer 1

Upon executing a command as $(command) allows the output of the command to replace itself.

When you say:

a=$(false)             # false fails; the output of false is stored in the variable a

the output produced by the command false is stored in the variable a. Moreover, the exit code is the same as produced by the command. help false would tell:

false: false
    Return an unsuccessful result.

    Exit Status:
    Always fails.

On the other hand, saying:

$ false                # Exit code: 1
$ a=""                 # Exit code: 0
$ echo $?              # Prints 0

causes the exit code for the assignment to a to be returned which is 0.

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EDIT:

Quoting from the manual:

If one of the expansions contained a command substitution, the exit status of the command is the exit status of the last command substitution performed.

Quoting from BASHFAQ/002:

How can I store the return value and/or output of a command in a variable?

...

output=$(command)

status=$?

The assignment to output has no effect on command's exit status, which is still in $?.