Bash中变量赋值命令替换的退出代码

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 using local within a function. Which is subtly different behavior than described in the the accepted answer, and the posted link here: http://mywiki.wooledge.org/BashFAQ/002

Take this bash script for example:

#!/bin/bash
function funWithLocalVar() {
    local output="$(echo "Doing some stuff.";exit 1)"
    local exitCode=$?
    echo "output: $output"
    echo "exitCode: $exitCode"
}

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

funWithLocalVar
funWithoutLocalVar

Here is the output of this:

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

Maybe nobody cares, but I did. It took me a minute to figure out why my status code was always 0 when it clearly wasn't sometimes. Not 100% clear why. But just knowing this helped.

Answer 3

I came across the same problem yesterday (Aug 29 2018).

In addition to local mentioned in Nick P.'s answer and @sevko's comment in the accepted answer , declare in global scope also has the same behavior.

Here's my Bash code:

#!/bin/bash

func1()
{
    ls file_not_existed
    local local_ret1=$?
    echo "local_ret1=$local_ret1"

    local local_var2=$(ls file_not_existed)
    local local_ret2=$?
    echo "local_ret2=$local_ret2"

    local local_var3
    local_var3=$(ls file_not_existed)
    local local_ret3=$?
    echo "local_ret3=$local_ret3"
}

func1

ls file_not_existed
global_ret1=$?
echo "global_ret1=$global_ret1"

declare global_var2=$(ls file_not_existed)
global_ret2=$?
echo "global_ret2=$global_ret2"

declare global_var3
global_var3=$(ls file_not_existed)
global_ret3=$?
echo "global_ret3=$global_ret3"

The output:

$ ./declare_local_command_substitution.sh 2>/dev/null 
local_ret1=2
local_ret2=0
local_ret3=2
global_ret1=2
global_ret2=0
global_ret3=2

Note the values of local_ret2 and global_ret2 in the output above. The exit codes are overwritten by local and declare .

My Bash version:

$ echo $BASH_VERSION 
4.4.19(1)-release

Answer 4

(not an answer to original question but too long for comment)

Note that export A=$(false); echo $? export A=$(false); echo $? outputs 0! Apparently the rules quoted in devnull's answer no longer apply. To add a bit of context to that quote (emphasis mine):

3.7.1 Simple Command Expansion

...

If there is a command name left after expansion, execution proceeds as described below . Otherwise , the command exits. 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. If there were no command substitutions, the command exits with a status of zero.

3.7.2 Command Search and Execution [ — this is the "below" case ]

IIUC the manual describes var=foo as special case of var=foo command... syntax (pretty confusing!). The "exit status of the last command substitution" rule only applies to the no-command case.

While it's tempting to think of export var=foo as a "modified assignment syntax", it isn't — export is a builtin command (that just happens to take assignment-like args).

=> If you want to export a var AND capture command substitution status, do it in 2 stages:

A=$(false)
# ... check $?
export A

This way also works in set -e mode — exits immediately if the command substitution return non-0.

Answer 5

As others have said, the exit code of the command substitution is the exit code of the substituted command, so

FOO=$(false)
echo $?
---
1

However, unexpectedly, adding export to the beginning of that produces a different result:

export FOO=$(false)
echo $?
---
0

This is because, while the substituted command false fails, the export command succeeds, and that is the exit code returned by the statement.