Determine if a function exists in bash

Question

Like this: [[ $(type -t foo) == function ]] && echo "Foo exists"

The built-in type command will tell you whether something is a function, built-in function, external command, or just not defined.

Additional examples:

$ LC_ALL=C type foo
bash: type: foo: not found

$ LC_ALL=C type ls
ls is aliased to `ls --color=auto'

$ which type

$ LC_ALL=C type type
type is a shell builtin

$ LC_ALL=C type -t rvm
function

$ if [ -n "$(LC_ALL=C type -t rvm)" ] && [ "$(LC_ALL=C type -t rvm)" = function ]; then echo rvm is a function; else echo rvm is NOT a function; fi
rvm is a function

The builtin bash command declare has an option -F that displays all defined function names. If given name arguments, it will display which of those functions are exist, and if all do it will set status accordingly:

$ fn_exists() { declare -F "$1" > /dev/null; }

$ unset f
$ fn_exists f && echo yes || echo no
no

$ f() { return; }
$ fn_exist f && echo yes || echo no
yes

If declare is 10x faster than test, this would seem the obvious answer.

Edit: Below, the -f option is superfluous with BASH, feel free to leave it out. Personally, I have trouble remembering which option does which, so I just use both. -f shows functions, and -F shows function names.

#!/bin/sh

function_exists() {
    declare -f -F $1 > /dev/null
    return $?
}

function_exists function_name && echo Exists || echo No such function

The "-F" option to declare causes it to only return the name of the found function, rather than the entire contents.

There shouldn't be any measurable performance penalty for using /dev/null, and if it worries you that much:

fname=`declare -f -F $1`
[ -n "$fname" ]    && echo Declare -f says $fname exists || echo Declare -f says $1 does not exist

Or combine the two, for your own pointless enjoyment. They both work.

fname=`declare -f -F $1`
errorlevel=$?
(( ! errorlevel )) && echo Errorlevel says $1 exists     || echo Errorlevel says $1 does not exist
[ -n "$fname" ]    && echo Declare -f says $fname exists || echo Declare -f says $1 does not exist

Borrowing from other solutions and comments, I came up with this:

fn_exists() {
  # appended double quote is an ugly trick to make sure we do get a string -- if $1 is not a known command, type does not output anything
  [ `type -t $1`"" == 'function' ]
}

Used as ...

if ! fn_exists $FN; then
    echo "Hey, $FN does not exist ! Duh."
    exit 2
fi

It checks if the given argument is a function, and avoids redirections and other grepping.

Dredging up an old post ... but I recently had use of this and tested both alternatives described with :

test_declare () {
    a () { echo 'a' ;}

    declare -f a > /dev/null
}

test_type () {
    a () { echo 'a' ;}
    type a | grep -q 'is a function'
}

echo 'declare'
time for i in $(seq 1 1000); do test_declare; done
echo 'type'
time for i in $(seq 1 100); do test_type; done

this generated :

real    0m0.064s
user    0m0.040s
sys     0m0.020s
type

real    0m2.769s
user    0m1.620s
sys     0m1.130s

declare is a helluvalot faster !