这告诉你它是否存在，但不是说它是一个函数

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

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。

像这样:[[$(type -t foo) == function]] && echo " foo存在"

内置类型命令将告诉您某个东西是函数、内置函数、外部命令还是没有定义。

附加的例子:

$ 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

它可以归结为使用“declare”来检查输出或退出代码。

输出方式:

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

用法:

isFunction some_name && echo yes || echo no

然而，如果内存有用，重定向到null比输出替换更快(说到这里，可怕的过时的' cmd '方法应该被摒弃，取而代之的是$(cmd)。)由于declare在找到/未找到时返回true/false，并且函数返回函数中最后一个命令的退出码，因此通常不需要显式返回，并且由于检查错误代码比检查字符串值(甚至是空字符串)更快:

退出状态样式:

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

这可能是你能得到的最简洁和最温和的说法了。

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
$

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