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 $? 

It outputs 1, which let me think that variable assignment doesn't sweep or produce new error code upon the last one. But when I tried this:

false; a=""; echo $? 

It outputs 0, obviously this is what a="" returns and it override 1 returned by false.

I want to know why this happens, is there any particularity in variable assignment that differs from other normal commands? Or just be cause a=$(false) is considered to be a single command and only command substitution part make sense?

-- UPDATE --

Thanks everyone, from the answers and comments I got the point "When you assign a variable using command substitution, the exit status is the status of the command." (by @Barmar), this explanation is excellently clear and easy to understand, but speak doesn't precise enough for programmers, I want to see the reference of this point from authorities such as TLDP or GNU man page, please help me find it out, thanks again!

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.

---

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 $?.

Answer 2

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.