借鉴其他解决方案和评论，我得出了以下结论:

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' ]
}

用作……

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

它检查给定的参数是否是一个函数，并避免重定向和其他grepping。

内置bash命令声明有一个选项-F，显示所有定义的函数名。如果给出了name参数，它将显示这些函数中哪些存在，如果都存在，它将相应地设置状态:

$ 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

测试不同的解决方案:

#!/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

例如:输出

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

所以declare -F似乎是最好的解决方案。

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

更新

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

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

如果declare比test快10倍，这似乎是显而易见的答案。

编辑:下面，-f选项在BASH中是多余的，可以省略它。就我个人而言，我很难记住哪个选项能做哪个，所以我两种都用。-f显示函数，-f显示函数名。

#!/bin/sh

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

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

要声明的“-F”选项导致它只返回找到的函数的名称，而不是整个内容。

使用/dev/null不应该有任何可衡量的性能损失，如果您如此担心:

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

或者把两者结合起来，只为你自己毫无意义的享受。它们都有效。

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

借鉴其他解决方案和评论，我得出了以下结论:

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' ]
}

用作……

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

它检查给定的参数是否是一个函数，并避免重定向和其他grepping。