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(

Answer 1

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 2

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!

Answer 3

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 4

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

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

Answer 5

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 6

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