bash: passing entire command (with arguments) to a function

Question

I am essentially trying to implement a function which asserts the failure (non-zero exit code) of another command, and prints a message when it fails.

Here is my function:

function assert_fail () {
    COMMAND=$@
    if [ `$COMMAND; echo $?` -ne 0 ]; then 
        echo "$COMMAND failed as expected."
    else
        echo "$COMMAND didn't fail"
    fi
}

# This works as expected
assert_fail rm nonexistent

# This works too
assert_fail rm nonexistent nonexistent2

# This one doesn't work
assert_fail rm -f nonexixtent

As soon as I add options to the command, it doesn't work. Here is the output of the above:

rm: cannot remove `nonexistent': No such file or directory
rm nonexistent failed as expected.
rm: cannot remove `nonexistent': No such file or directory
rm: cannot remove `nonexistent2': No such file or directory
rm nonexistent nonexistent2 failed as expected.
rm -f nonexistent didn't fail

I have tried putting double quotes around the commands, to no avail. I would expect the third invocation in the above to produce similar output to the other two.

I appreciate any/all help!

You should use COMMAND="$@" (quote the arguments) and if $COMMAND; then (test the exit status directly, instead of echoing and capturing the exit status for explicit comparison). — chepner
– chepner, Commented Oct 15, 2012 at 19:13

Gordon Davisson · Accepted Answer · 2012-10-16 03:14:03Z

13

@rici correctly pointed out the issue you're seeing, but there are a couple of real problems with your wrapper function. First, it doesn't correctly preserve spaces (and some other funny characters) in arguments. COMMAND=$@ (or COMMAND="$@") merges all of the arguments into a single string, losing the distinction between spaces between arguments and spaces within arguments. To keep them straight, either use "$@" directly without storing it in a variable, or store it as an array (COMMAND=("$@"), then execute it as "${COMMAND[@]}"). Second, if the command prints anything to stdout, it'll wreak havoc with your exit status check; just test it directly, as @chepner said. Here's my suggested rewrite:

function assert_fail () {
    if "$@"; then 
        echo "$* didn't fail"
    else
        echo "$* failed as expected."
    fi
}

Note that the way I did the echo commands does lose the distinction of spaces within arguments. If that's a problem, replace the echo commands with this:

printf "%q " "$@"
echo "didn't fail"

and

printf "%q " "$@"
echo "failed as expected."

answered Oct 16, 2012 at 3:14

Gordon Davisson

128k16 gold badges141 silver badges168 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Justin Lewis Over a year ago

You're right, that would have caused me problems later on. Thanks. +1

rici · Accepted Answer · 2012-10-15 15:36:52Z

5

rm -f never fails on non-existent files. It has nothing to do with your wrapper. See man rm:

OPTIONS
       -f, --force
              ignore nonexistent files, never prompt

answered Oct 15, 2012 at 15:36

rici

243k30 gold badges263 silver badges364 bronze badges

Collectives™ on Stack Overflow

bash: passing entire command (with arguments) to a function

2 Answers 2

1 Comment

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Related