Bash – Why Parentheses Return Exit Status but Not Braces

bashshell-script

I understand that parentheses cause the commands to be run in a subshell and braces cause the commands to be grouped together but not in a subshell.

When I run this with parentheses:

no_func || (
    echo "there is nothing"
    exit 1
)

echo $?

This returns the exit status:

/Users/myname/bin/ex5: line 34: n_func: command not found
there is nothing
1

But when I use braces:

no_func || {
    echo "there is nothing"
    exit 1
}

echo $?

This doesn't return the exit status.

/Users/myname/bin/ex5: line 34: no_func: command not found
there is nothing

But why one returns the exit status and another doesn't?

Best Answer

Look at the command execution trace (set -x). With braces:

+ no_func
./a: line 3: no_func: command not found
+ echo 'there is nothing'
there is nothing
+ exit 1

exit exits the (sub)shell. Since braces don't create a subshell, exit exits the main shell process, so it never reaches the point where it would run echo $?.

Related Solutions

Shell – How to keep last exit status after test

$_ will work in (at least) interactive dash, bash, zsh, ksh (though apparently not in a conditional statement as requested) and mksh shells. Of those - to my knowledge - only bash and zsh will also populate it in a scripted shell. It is not a POSIX parameter - but is fairly portable to any modern, interactive shell.

For a more portable solution you can do:

command -p sudo ...
eval '[ "$?" = 127 ] || exit '"$?"

Which basically allows you to expand the initial value for $? into the tail of the script before ever even testing its value at its head.

But anyway, since you appear to be testing whether or not the command sudo can be found in the shell's builtin -p portable path string with command, I would think you could go at it a litte more directly. Also, just to be clear, command won't test for the location of any arguments to sudo - so it is only sudo - and nothing it invokes - which is relevant to that return value.

And so anyway, if that is what you're trying to do:

command -pv sudo >/dev/null || handle_it
command -p  sudo something or another

...would work just fine as a test without any chance of errors in the command sudo runs returning in such a way that might skew the results of your test.

Bash – How to catch and handle nonzero exit status within a Bash function

Under set -e, the non-existence of failfailfail causes the whole script to exit (or the subshell, if the function is executed in a subshell).

If you don't need to modify the shell state from the function, you can run it in a subshell.

myfunc() (
    set -e
    ls
    failfailfail
    uptime
)

Another approach in bash is to set an ERR trap to execute return. If you want to make it a local setting, you need to restore the old trap value, which is a little cumbersome.

myfunc() {
    local old_ERR_trap=$(trap -p ERR)
    if [[ -z $old_ERR_trap ]]; then old_ERR_trap="trap - ERR"; fi
    trap 'local ret=$?; eval "$old_ERR_trap"; return $ret' ERR
    ls
    failfailfail
    uptime
}

Best Answer

Related Solutions

Shell – How to keep last exit status after test

Bash – How to catch and handle nonzero exit status within a Bash function

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