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

Testing different solutions:

#!/bin/bash

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

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

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

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

funcs=(test_declare test_declare2 test_type test_type2)

test () {
    for i in $(seq 1 1000); do $1; done
}

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

for j in 1 2 3; do

    for func in ${funcs[@]}; do
        echo $func $post
        time test $func
        echo exit code $?; echo
    done

    case $j in
    1)  unset -f f
        post='(f unset)'
        ;;
    2)  f='string'
        post='(f is string)'
        ;;
    esac
done

outputs e.g.:

test_declare (f is function)

real 0m0,055s user 0m0,041s sys 0m0,004s exit code 0

test_declare2 (f is function)

real 0m0,042s user 0m0,022s sys 0m0,017s exit code 0

test_type (f is function)

real 0m2,200s user 0m1,619s sys 0m1,008s exit code 0

test_type2 (f is function)

real 0m0,746s user 0m0,534s sys 0m0,237s exit code 0

test_declare (f unset)

real 0m0,040s user 0m0,029s sys 0m0,010s exit code 1

test_declare2 (f unset)

real 0m0,038s user 0m0,038s sys 0m0,000s exit code 1

test_type (f unset)

real 0m2,438s user 0m1,678s sys 0m1,045s exit code 1

test_type2 (f unset)

real 0m0,805s user 0m0,541s sys 0m0,274s exit code 1

test_declare (f is string)

real 0m0,043s user 0m0,034s sys 0m0,007s exit code 1

test_declare2 (f is string)

real 0m0,039s user 0m0,035s sys 0m0,003s exit code 1

test_type (f is string)

real 0m2,394s user 0m1,679s sys 0m1,035s exit code 1

test_type2 (f is string)

real 0m0,851s user 0m0,554s sys 0m0,294s exit code 1

So declare -F f seems to be the best solution.