Determine if a function exists in bash

Question

Currently I'm doing some unit tests which are executed from bash. Unit tests are initialized, executed and cleaned up in a bash script. This script usualy contains an init(), execute() and cleanup() functions. But they are not mandatory. I'd like to test if they are or are not defined.

I did this previously by greping and seding the source, but it seemed wrong. Is there a more elegant way to do this?

Edit: The following sniplet works like a charm:

fn_exists()
{
    LC_ALL=C type $1 | grep -q 'shell function'
}

Answer 1

I think you're looking for the 'type' command. It'll tell you whether something is a function, built-in function, external command, or just not defined. Example:

$ 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

Answer 2

$ g() { return; }
$ declare -f g > /dev/null; echo $?
0
$ declare -f j > /dev/null; echo $?
1

Answer 3

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

Answer 4

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.

Answer 5

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 !

Answer 6

It boils down to using 'declare' to either check the output or exit code.

Output style:

isFunction() { [[ "$(declare -Ff "$1")" ]]; }

Usage:

isFunction some_name && echo yes || echo no

However, if memory serves, redirecting to null is faster than output substitution (speaking of, the awful and out-dated `cmd` method should be banished and $(cmd) used instead.) And since declare returns true/false if found/not found, and functions return the exit code of the last command in the function so an explicit return is usually not necessary, and since checking the error code is faster than checking a string value (even a null string):

Exit status style:

isFunction() { declare -Ff "$1" >/dev/null; }

That's probably about as succinct and benign as you can get.

Answer 7

Testing speed of different solutions

#!/bin/bash

f () {
echo 'This is a test function.'
echo 'This has more than one command.'
return 0
}

test_declare () {
    declare -f f > /dev/null
}

test_declare2 () {
    declare -F f > /dev/null
}

test_type () {
    type -t f | grep -q 'function'
}

test_type2 () {
    local var=$(type -t f)
    [[ "${var-}" = function ]]
}

post=
for j in 1 2; do
echo
echo 'declare -f' $post
time for i in $(seq 1 1000); do test_declare; done
echo
echo 'declare -F' $post
time for i in $(seq 1 1000); do test_declare2; done
echo
echo 'type with grep' $post
time for i in $(seq 1 1000); do test_type; done
echo
echo 'type with var' $post
time for i in $(seq 1 1000); do test_type2; done
unset -f f
post='(f unset)'
done

outputs e.g.:

declare -f

real 0m0.037s user 0m0.024s sys 0m0.012s

declare -F

real 0m0.030s user 0m0.020s sys 0m0.008s

type with grep

real 0m1.772s user 0m0.084s sys 0m0.340s

type with var

real 0m0.770s user 0m0.096s sys 0m0.160s

declare -f (f unset)

real 0m0.031s user 0m0.028s sys 0m0.000s

declare -F (f unset)

real 0m0.031s user 0m0.020s sys 0m0.008s

type with grep (f unset)

real 0m1.859s user 0m0.100s sys 0m0.348s

type with var (f unset)

real 0m0.683s user 0m0.092s sys 0m0.160s

So declare -F f && echo function f exists. || echo function f does not exist. seems to be the best solution.

Answer 8

fn_exists()
{
   [[ $(type -t $1) == function ]] && return 0
}

update

isFunc () 
{ 
    [[ $(type -t $1) == function ]]
}

$ isFunc isFunc
$ echo $?
0
$ isFunc dfgjhgljhk
$ echo $?
1
$ isFunc psgrep && echo yay
yay
$

Answer 9

This tells you if it exists, but not that it's a function

fn_exists()
{
  type $1 >/dev/null 2>&1;
}

Answer 10

I particularly liked solution from Grégory Joseph

But I've modified it a little bit to overcome "double quote ugly trick":

function is_executable()
{
    typeset TYPE_RESULT="`type -t $1`"

    if [ "$TYPE_RESULT" == 'function' ]; then
        return 0
    else
        return 1
    fi
}

Answer 11

From my comment on another answer (which I keep missing when I come back to this page)

$ fn_exists() { test x$(type -t $1) = xfunction; }
$ fn_exists func1 && echo yes || echo no
no
$ func1() { echo hi from func1; }
$ func1
hi from func1
$ fn_exists func1 && echo yes || echo no
yes

Answer 12

I would improve it to:

fn_exists()
{
    type $1 2>/dev/null | grep -q 'is a function'
}

And use it like this:

fn_exists test_function
if [ $? -eq 0 ]; then
    echo 'Function exists!'
else
    echo 'Function does not exist...'
fi

Answer 13

It is possible to use 'type' without any external commands, but you have to call it twice, so it still ends up about twice as slow as the 'declare' version:

test_function () {
        ! type -f $1 >/dev/null 2>&1 && type -t $1 >/dev/null 2>&1
}

Plus this doesn't work in POSIX sh, so it's totally worthless except as trivia!