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?

Adam Bryzak · Accepted Answer · 2011-06-16T00:24:57.350

66

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.

edited Jun 16 '11 at 00:24

answered Jun 16 '11 at 00:19

Adam Bryzak

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4

A benefit of this approach over using the `which` utility is that it will avoid an extra process being launched. – Adam Bryzak Jun 16 '11 at 00:19
And POSIX http://pubs.opengroup.org/onlinepubs/9699919799/utilities/command.html , awesome. – Ciro Santilli OurBigBook.com Aug 05 '16 at 12:46

score 15 · Answer 2 · answered Jun 16 '11 at 00:02

15

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.

answered Jun 16 '11 at 00:02

Gustavo Giráldez

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Invoking program when a bash function has the same name

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