blocks|key|1758218|text|像这样：[[+$(type+-t+foo)+==+function+]]+&&+echo+"Foo+exists"|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1758219|内置的type命令将告诉您某些东西是函数、内置函数、外部命令还是未定义的。|1758220|其他示例：|1758221|$+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|code-block|syntax|javascript|1758222|entityMap^0|4|1H|0|3|4|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|T|8|@$9|U|A|V|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|W|8|@]|D|@]|E|$]]|$1|J|3|K|5|L|7|X|8|@]|D|@]|E|$M|N]]|$1|O|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|P|$]]

Like this: <code>[[ $(type -t foo) == function ]] &amp;&amp; echo &quot;Foo exists&quot;</code>
The built-in <code>type</code> command will tell you whether something is a function, built-in function, external command, or just not defined.
Additional examples:
<pre class="lang-none prettyprint-override"><code>$ 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 &quot;$(LC_ALL=C type -t rvm)&quot; ] &amp;&amp; [ &quot;$(LC_ALL=C type -t rvm)&quot; = function ]; then echo rvm is a function; else echo rvm is NOT a function; fi
rvm is a function
</code></pre>

blocks|key|1543332|text|内置bash命令declare有一个选项-F，用于显示所有已定义的函数名。如果给定了名称参数，它将显示这些函数中存在的函数，如果所有函数都存在，它将相应地设置状态：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1543333|$+fn_exists()+{+declare+-F+"$1"+>+/dev/null;+}

$+unset+f
$+fn_exists+f+&&+echo+yes+%7C%7C+echo+no
no

$+f()+{+return;+}
$+fn_exist+f+&&+echo+yes+%7C%7C+echo+no
yes|code-block|syntax|javascript|1543334|entityMap^0|8|7|K|2|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

The builtin bash command <code>declare</code> has an option <code>-F</code> that displays all defined function names. If given name arguments, it will display which of those functions are exist, and if all do it will set status accordingly:
<pre class="lang-sh prettyprint-override"><code>$ fn_exists() { declare -F &quot;$1&quot; &gt; /dev/null; }

$ unset f
$ fn_exists f &amp;&amp; echo yes || echo no
no

$ f() { return; }
$ fn_exist f &amp;&amp; echo yes || echo no
yes
</code></pre>

blocks|key|1543431|text|这会告诉您它是否存在，但不会告诉您它是一个函数|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1543432|fn_exists()
{
++type+$1+>/dev/null+2>&1;
}|code-block|syntax|javascript|1543433|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

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

<pre><code>fn_exists()
{
 type $1 &gt;/dev/null 2&gt;&amp;1;
}
</code></pre>

blocks|key|1543377|text|我会将其改进为：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1543378|fn_exists()
{
++++type+$1+2>/dev/null+%7C+grep+-q+'is+a+function'
}|code-block|syntax|javascript|1543379|并像这样使用它：|1543380|fn_exists+test_function
if+[+$?+-eq+0+];+then
++++echo+'Function+exists!'
else
++++echo+'Function+does+not+exist...'
fi|1543381|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

I would improve it to:

<pre><code>fn_exists()
{
 type $1 2&gt;/dev/null | grep -q 'is a function'
}
</code></pre>

And use it like this:

<pre><code>fn_exists test_function
if [ $? -eq 0 ]; then
 echo 'Function exists!'
else
 echo 'Function does not exist...'
fi
</code></pre>

blocks|key|1758424|text|翻出一个旧帖子..。但我最近使用了这一点，并测试了下面描述的两种替代方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1758425|test_declare+()+{
++++a+()+{+echo+'a'+;}

++++declare+-f+a+>+/dev/null
}

test_type+()+{
++++a+()+{+echo+'a'+;}
++++type+a+%7C+grep+-q+'is+a+function'
}

echo+'declare'
time+for+i+in+$(seq+1+1000);+do+test_declare;+done
echo+'type'
time+for+i+in+$(seq+1+100);+do+test_type;+done|code-block|syntax|javascript|1758426|这就产生了：|1758427|real++++0m0.064s
user++++0m0.040s
sys+++++0m0.020s
type

real++++0m2.769s
user++++0m1.620s
sys+++++0m1.130s|1758428|声明是一个地狱般的速度更快！|1758429|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Dredging up an old post ... but I recently had use of this and tested both alternatives described with : 

<pre><code>test_declare () {
 a () { echo 'a' ;}

 declare -f a &gt; /dev/null
}

test_type () {
 a () { echo 'a' ;}
 type a | grep -q 'is a function'
}

echo 'declare'
time for i in $(seq 1 1000); do test_declare; done
echo 'type'
time for i in $(seq 1 100); do test_type; done
</code></pre>

this generated : 

<pre><code>real 0m0.064s
user 0m0.040s
sys 0m0.020s
type

real 0m2.769s
user 0m1.620s
sys 0m1.130s
</code></pre>

declare is a helluvalot faster !

blocks|key|1758485|text|可以在没有任何外部命令的情况下使用'type‘，但你必须调用它两次，所以它最终仍然比'declare’版本慢两倍：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1758486|test_function+()+{
++++++++!+type+-f+$1+>/dev/null+2>&1+&&+type+-t+$1+>/dev/null+2>&1
}|code-block|syntax|javascript|1758487|另外，这在POSIX+sh中不起作用，所以除了作为琐事之外，它是完全没有价值的！|1758488|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

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:

<pre><code>test_function () {
 ! type -f $1 &gt;/dev/null 2&gt;&amp;1 &amp;&amp; type -t $1 &gt;/dev/null 2&gt;&amp;1
}
</code></pre>

Plus this doesn't work in POSIX sh, so it's totally worthless except as trivia!

blocks|key|1758526|text|借鉴其他解决方案和评论，我得出了以下结论：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1758527|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+$1`""+==+'function'+]
}|code-block|syntax|javascript|1758528|用作...|1758529|if+!+fn_exists+$FN;+then
++++echo+"Hey,+$FN+does+not+exist+!+Duh."
++++exit+2
fi|1758530|它检查给定的参数是否是函数，并避免重定向和其他返回值。|1758531|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Borrowing from other solutions and comments, I came up with this:

<pre><code>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' ]
}
</code></pre>

Used as ...

<pre><code>if ! fn_exists $FN; then
 echo "Hey, $FN does not exist ! Duh."
 exit 2
fi
</code></pre>

It checks if the given argument is a function, and avoids redirections and other grepping.

blocks|key|1758604|text|我特别喜欢Grégory+Joseph的解决方案|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1758605|但我对它做了一点修改，以克服“双引号丑陋的技巧”：|1758606|function+is_executable()
{
++++typeset+TYPE_RESULT="`type+-t+$1`"

++++if+[+"$TYPE_RESULT"+==+'function'+];+then
++++++++return+0
++++else
++++++++return+1
++++fi
}|code-block|syntax|javascript|1758607|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/9002012/216172^0|5|E|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|6|7|W|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|X|8|@]|9|@]|C|$I|J]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]]]

I particularly liked solution from <a href="https://stackoverflow.com/a/9002012/216172">Grégory Joseph</a>

But I've modified it a little bit to overcome "double quote ugly trick":

<pre><code>function is_executable()
{
 typeset TYPE_RESULT="`type -t $1`"

 if [ "$TYPE_RESULT" == 'function' ]; then
 return 0
 else
 return 1
 fi
}
</code></pre>

blocks|key|1543724|text|它归结为使用'declare‘来检查输出或退出代码。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1543725|输出样式：|1543726|isFunction()+{+[[+"$(declare+-Ff+"$1")"+]];+}|code-block|syntax|javascript|1543727|用法：|1543728|isFunction+some_name+&&+echo+yes+%7C%7C+echo+no|1543729|然而，如果内存有效，重定向到null比输出替换更快(说到这里，可怕和过时的cmd方法应该被驱逐，而使用$(cmd)。)而且，由于declare在找到/未找到时返回true/false，而函数返回函数中最后一个命令的退出代码，因此通常不需要显式返回，而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：|offset|length|style|CODE|1543730|退出状态样式：|1543731|isFunction()+{+declare+-Ff+"$1"+>/dev/null;+}|1543732|这可能是您所能得到的最简洁和最友好的方式了。|1543733|entityMap^0|0|0|0|0|0|11|3|0|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|11|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|12|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|13|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|14|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|15|8|@$O|16|P|17|Q|R]]|9|@]|A|$]]|$1|S|3|T|5|6|7|18|8|@]|9|@]|A|$]]|$1|U|3|V|5|F|7|19|8|@]|9|@]|A|$G|H]]|$1|W|3|X|5|6|7|1A|8|@]|9|@]|A|$]]|$1|Y|3|-4|5|6|7|1B|8|@]|9|@]|A|$]]]|Z|$]]

It boils down to using 'declare' to either check the output or exit code.

Output style:

<pre><code>isFunction() { [[ "$(declare -Ff "$1")" ]]; }
</code></pre>

Usage:

<pre><code>isFunction some_name &amp;&amp; echo yes || echo no
</code></pre>

However, if memory serves, redirecting to null is faster than output substitution (speaking of, the awful and out-dated `cmd` method should be banished and $(cmd) used instead.) And since declare returns true/false if found/not found, and functions return the exit code of the last command in the function so an explicit return is usually not necessary, and since checking the error code is faster than checking a string value (even a null string):

Exit status style:

<pre><code>isFunction() { declare -Ff "$1" &gt;/dev/null; }
</code></pre>

That's probably about as succinct and benign as you can get.

blocks|key|1543775|text|来自我对另一个答案的评论(当我回到这个页面时，我一直遗漏了这个答案)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1543776|$+fn_exists()+{+test+x$(type+-t+$1)+=+xfunction;+}
$+fn_exists+func1+&&+echo+yes+%7C%7C+echo+no
no
$+func1()+{+echo+hi+from+func1;+}
$+func1
hi+from+func1
$+fn_exists+func1+&&+echo+yes+%7C%7C+echo+no
yes|code-block|syntax|javascript|1543777|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

From my comment on another answer (which I keep missing when I come back to this page)

<pre><code>$ fn_exists() { test x$(type -t $1) = xfunction; }
$ fn_exists func1 &amp;&amp; echo yes || echo no
no
$ func1() { echo hi from func1; }
$ func1
hi from func1
$ fn_exists func1 &amp;&amp; echo yes || echo no
yes
</code></pre>

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(), execute() and cleanup() functions. But they are not mandatory. I'd like to test if they are or are not defined.

I did this previously by greping and seding the source, but it seemed wrong. Is there a more elegant way to do this?

Edit: The following sniplet works like a charm:

<pre><code>fn_exists()
{
 LC_ALL=C type $1 | grep -q 'shell function'
}
</code></pre>

Determine if a function exists in bash

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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目前我正在做一些单元测试，这些单元测试是从bash执行的。单元测试是在bash脚本中初始化、执行和清理的。此脚本通常包含init()、execute()和cleanup()函数。但它们不是强制性的。我想测试一下是否定义了它们。我之前通过贪婪和seding源代码做到了这一点，但这似乎是错误的。有没有更优雅的方式来做这件事...

问判断bash中是否存在函数
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