How to pass a parameter in a command without run the parameter content?

Good days. I have a little problem. I have this function in bash script function denyCheck(){ file=$(readlink -f $(which $(printf "%s\n" "$1"))) if [[ -x $file ]];then return false else return true fi } denyCheck "firefox" This function I pass her a string, what is a comman, and this resolv the orginal path where is the command (by example: /usr/lib/firefox/firefox.sh) and if the file is executable return true else false. But the problem is...The parameter (in the example "firefox") run it as command and open the browser firefox. How can I do for that the parameter not run it? Thank you very much.