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

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