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242

answers:

3

Let's imagine I have a bash script, where I call this:

bash -c "some_command"
do something with code of some_command here

Is it possible to obtain the code of some_command? I'm not executing some_command directly in the shell running the script because I don't want to alter it's environment.

+2  A: 

$? will contain the return code of some_command just as usual.

Edit: Of course it might also contain a code from bash, in case something went wrong before your command could even be executed (wrong filename, whatnot)

Matti Virkkunen
+1  A: 

You can use the $? variable, check out the bash documentation for this, it stores the exit status of the last command.

Also, you might want to check out the bracket-style command blocks of bash (e.g. comm1 && (comm2 || comm3) && comm4), they are always executed in a subshell thus not altering the current environment, and are more powerful as well!

EDIT: For instance, when using ()-style blocks as compared to bash -c 'command', you don't have to worry about escaping any argument strings with spaces, or any other special shell syntax. You directly use the shell syntax, it's a normal part of the rest of the code.

Paggas
In other words, `(some_command)` instead of `bash -c "some_command"`
Dennis Williamson
+1  A: 

Here's an illustration of $? and the parenthesis subshell mentioned by Paggas and Matti:

$ (exit a); echo $?
-bash: exit: a: numeric argument required
255
$ (exit 33); echo $?
33

In the first case, the code is a Bash error and in the second case it's the exit code of exit.

Dennis Williamson