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up vote 25 down vote favorite

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!

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up vote 37 down vote accepted

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 $?.

share|improve this answer
    
Very good answer :)! Can you please help me find out the reference of the exit code is the same as produced by the command, I spend a lot of time but did not get the result :( – Reorx Nov 23 '13 at 3:25
    
@Reorx Added a couple of references in the edit above. – devnull Nov 23 '13 at 3:31
    
Thanks, that's clear enough :D – Reorx Nov 23 '13 at 3:39
7  
Keep in mind that this doesn't apply to local variables (eg, local output=$(command)), because local is a command itself and its exit-code will overwrite that of the assigned function. – sevko Jul 19 '14 at 2:45
3  
Regarding @sevko & @qodeninja 's comments about locals, this can easily be circumvented by initialization after declaration unless the variable is also readonly: local my_var; my_var=$(command); local status=$? – Adrian Günter Jul 26 '15 at 1:10
up vote -2 down vote

You are using the last return variable

$?

gives the return value of the last function or program executed. Executing false will end in 0, but assigning something to false does not return, so the 1 you are getting is actually the return code of the program PREVIOUSLY executed.

share|improve this answer
    
No, a=$(false) does return 1 because of the false. – Kevin Nov 23 '13 at 3:00
3  
When you assign a variable using command substitution, the exit status is the status of the command. – Barmar Nov 23 '13 at 3:10

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