Bash – Running Executable in PATH with Same Name as Existing Function

bashfunctionpath

Sometimes I define a function that shadows an executable and tweaks its arguments or output. So the function has the same name as the executable, and I need a way how to run the executable from the function without calling the function recursively. For example, to automatically run the output of fossil diff through colordiff and less -R I use:

function fossil () {
    local EX=$(which fossil)
    if [ -z "$EX" ] ; then
        echo "Unable to find 'fossil' executable." >&2
        return 1
    fi
    if [ -t 1 ] && [ "$1" == "diff" ] ; then
        "$EX" "$@" | colordiff | less -R
        return
    fi
    "$EX" "$@"
}

If I'd be sure about the location of the executable, the I could simply type /usr/bin/fossil. Bash recognizes that / means the command it's an executable, not a function. But since I don't know the exact location, I have to resort to calling which and checking the result. Is there a simpler way?

Best Answer

Use the command shell builtin:

bash-4.2$ function date() { echo 'at the end of days...'; }

bash-4.2$ date
at the end of days...

bash-4.2$ command date
Mon Jan 21 16:24:33 EET 2013

bash-4.2$ help command
command: command [-pVv] command [arg ...]
    Execute a simple command or display information about commands.

    Runs COMMAND with ARGS suppressing  shell function lookup, or display
    information about the specified COMMANDs.  Can be used to invoke commands
    on disk when a function with the same name exists.
Related Question