[SOLVED] How to pass command with its arguments as a single argument to script?

Issue

I have a script I use for building my kernel, and in it I have a function I want to use to catch errors in my build commands, but I can't figure out how to pass multiple arguments as one to get it working properly. The syntax is as follows:

run_build() {
cmd=$($1)
status=${PIPESTATUS[0]}
$*
if ($status != 0); then
    echo "$cmd failed!"
    exit 1
else
    echo "$cmd succeeded."
fi
}

The problem is, when I try to pass a make command with its arguments, the make command is being executed with no arguments. I tried to use the "$*" syntax to call all arguments but it doesn't seem to work. An example command would be

make O=out -j8 zImage

I need the next 3 arguments to be included when calling the line from the script. Anyone have ideas?

Solution

First point, you don't want to pass the entire command as a single argument: I'm trying to put a command in a variable, but the complex cases always fail!

This is the main error: cmd=$($1) -- the $() syntax invokes Command Substitution that executes the first argument. Just use cmd=$1 to store the first parameter into the "cmd" variable.

Additional errors:

• if ($status != 0); then -- that syntax is incorrect: use if ((status != 0)); then for proper arithmetic evaluation.
• $* -- to execute the positional parameters correctly, use "$@" (with the quotes) -- that form will maintain any arguments that contain whitespace as intended.

Answered By - glenn jackman

Answer Checked By - Marie Seifert (WPSolving Admin)