Invoking program when a bash function has the same name

Question

I have the following function in my bash script:

make() {     cd Python-3.2     make } 

When make is called within this script, this function is invoked, which recurses. The call to make inside the function should actually invoke the external make utility. Other than renaming my make function, what's the cleanest way to achieve this?

Answer 1

You can use the command built-in to suppress shell function lookups.

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.      Options:       -p    use a default value for PATH that is guaranteed to find all of         the standard utilities       -v    print a description of COMMAND similar to the `type' builtin       -V    print a more verbose description of each COMMAND      Exit Status:     Returns exit status of COMMAND, or failure if COMMAND is not found. 

Answer 2

Use the full path to the program. E.g. /usr/bin/make.

If you don't know the full path, you can use the which utility, like:

$(which make) 

That will find the full path and execute make.