blocks|key|1314624|text|#include+<iostream>
using+namespace+std;

int+f1()+{
++++return+1;
}

int+f2()+{
++++return+2;
}

typedef+int+(*fptr)();


fptr+f(+char+c+)+{
++++if+(+c+==+'1'+)+{
++++++++return+f1;
++++}
++++else+{
++++++++return+f2;
++++}
}

int+main()+{
++++char+c+=+'1';
++++fptr+fp+=+f(+c+);
++++cout+<<+fp()+<<+endl;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1314625|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>#include &lt;iostream&gt;
using namespace std;

int f1() {
 return 1;
}

int f2() {
 return 2;
}

typedef int (*fptr)();


fptr f( char c ) {
 if ( c == '1' ) {
 return f1;
 }
 else {
 return f2;
 }
}

int main() {
 char c = '1';
 fptr fp = f( c );
 cout &lt;&lt; fp() &lt;&lt; endl;
}
</code></pre>

blocks|key|1314747|text|int+f(char)+{
++++return+0;
}

int+(*return_f())(char)+{
++++return+f;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1314748|不，说真的，请使用typedef+:)|unstyled|1314749|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>int f(char) {
 return 0;
}

int (*return_f())(char) {
 return f;
}
</code></pre>

No, seriously, use a typedef :)

blocks|key|1529688|text|为函数签名创建一个typedef：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1529689|typedef+void+(*+FuncSig)(int+param);|code-block|syntax|javascript|1529690|然后将您的函数声明为返回FuncSig：|1529691|FuncSig+GetFunction();|1529692|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|$]]

Create a typedef for the function signature:

<pre><code>typedef void (* FuncSig)(int param);
</code></pre>

Then declare your function as returning FuncSig:

<pre><code>FuncSig GetFunction();
</code></pre>

blocks|key|1314922|text|在C%2B%2B11中，您可以使用尾随返回类型来简化语法，例如，假设一个函数：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1314923|int+c(int+d)+{+return+d+*+2;+}|code-block|syntax|javascript|1314924|这可以从一个函数返回(需要使用双精度表示)：|1314925|int+(*foo(double+e))(int)
{
++++e;
++++return+c;
}|1314926|使用尾随返回类型，这将变得更容易阅读：|1314927|auto+foo2(double+e)+->+int(*)(int)
{
++++e;
++++return+c;
}|1314928|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

In C++11 you can use trailing return types to simplify the syntax, e.g. assuming a function: 

<pre><code>int c(int d) { return d * 2; }
</code></pre>

This can be returned from a function (that takes a double to show that):

<pre><code>int (*foo(double e))(int)
{
 e;
 return c;
}
</code></pre>

Using a trailing return type, this becomes a bit easier to read:

<pre><code>auto foo2(double e) -&gt; int(*)(int)
{
 e;
 return c;
}
</code></pre>

blocks|key|1314846|text|下面是如何在不使用typedef的情况下完成此操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1314847|int+c(){+return+0;+}

int+(*+foo+(void))(){++//compiles
return+c;
}|code-block|syntax|javascript|1314848|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|$]]

Here is how to do it without using a typedef:

<pre><code>int c(){ return 0; }

int (* foo (void))(){ //compiles
return c;
}
</code></pre>

blocks|key|1529752|text|typedef+void+(*voidFn)();

void+foo()
{
}

voidFn+goo(char+c)
{
++++if+(c+==+'f')+{
++++++++return+foo;
++++}
++++else+{
++++++++//..
++++}
++++//+..
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1529753|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>typedef void (*voidFn)();

void foo()
{
}

voidFn goo(char c)
{
 if (c == 'f') {
 return foo;
 }
 else {
 //..
 }
 // ..
}
</code></pre>

blocks|key|1529656|text|这是显示函数指针返回的代码。首先需要定义返回的“函数签名”：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1529657|int+returnsOne()+{
+++++return+1;
}

typedef+int(*fp)();

fp+returnsFPtoReturnsOne()+{
++++&returnsOne;
}|code-block|syntax|javascript|1529658|在您的特定情况下：|1529659|fp+getFunctionFor(char+code)+{
++++switch+(code)+{
++++++++case+'a':+return+&functionA;
++++++++case+'b':+return+&functionB;
++++}
++++return+NULL;
}|1529660|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|$]]

This is the code to show return of a function pointer. You need to define the "function signature" to return first:

<pre><code>int returnsOne() {
 return 1;
}

typedef int(*fp)();

fp returnsFPtoReturnsOne() {
 &amp;returnsOne;
}
</code></pre>

In your specific case:

<pre><code>fp getFunctionFor(char code) {
 switch (code) {
 case 'a': return &amp;functionA;
 case 'b': return &amp;functionB;
 }
 return NULL;
}
</code></pre>

blocks|key|1314817|text|请查看此网站-+http://cdecl.org|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1314818|帮助您将英语声明转换为C语言声明并返回！|1314819|很酷的东西！|1314820|此链接解码erikallen答案中的示例。int+(*return_f())(char)|1314821|entityMap|0|LINK|mutability|MUTABLE|url|http://cdecl.org/|1|http://cdecl.ridiculousfish.com/?q=int%2520%2528%252areturn_f%2528%2529%2529%2528char%2529^0|8|G|0|0|0|0|L|N|1|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@$A|U|B|V|1|W]]|C|$]]|$1|D|3|E|5|6|7|X|8|@]|9|@]|C|$]]|$1|F|3|G|5|6|7|Y|8|@]|9|@]|C|$]]|$1|H|3|I|5|6|7|Z|8|@]|9|@$A|10|B|11|1|12]]|C|$]]|$1|J|3|-4|5|6|7|13|8|@]|9|@]|C|$]]]|K|$L|$5|M|N|O|C|$P|Q]]|R|$5|M|N|O|C|$P|S]]]]

Check out this site - <a href="http://cdecl.org" rel="nofollow">http://cdecl.org</a>

Helps you convert english to C declarations and back!

Cool Stuff!

This link decodes the example in erikallen's answer.
<a href="http://cdecl.ridiculousfish.com/?q=int%20%28%2areturn_f%28%29%29%28char%29" rel="nofollow">int (*return_f())(char)</a>

blocks|key|1314643|text|最简单的方法是定义所需的指向函数的指针类型，然后使用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1314644|typedef+void+(*fnptr_t)(int,+int);
fptr_t+myfunc(char+*)+{+....|code-block|syntax|javascript|1314645|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|$]]

Easiest way is to typedef the pointer-to-function type you want, and then use that

<pre><code>typedef void (*fnptr_t)(int, int);
fptr_t myfunc(char *) { ....
</code></pre>

blocks|key|1529839|text|我更喜欢返回对象并调用操作符()。这样，你的函数就可以返回一个接口，所有的类都可以继承这个接口。也就是说，如果您使用的是C%2B%2B而不是C。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1529840|然后，您可以使用参数化因子方法根据您的输入返回对象。|1529841|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

I prefer returning objects and call the operator(). This way your function can return an interface and all classes can inherit from this. That is, if you're using C++ and not C.

Then you can use the parametrized factor method to return the objects based on your input.

blocks|key|1314601|text|我假设C在这里(没有对象)+:)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1314602|//+Type+of+function+which+takes+a+char+and+returns+an+int:
typedef+int+(*Func)(char+a);

//+An+example+of+the+function+you're+trying+to+return+and+which+does+something
//+with+char:
int+exampleFunc(char+a)
{
++++return+(int)(a+%2B+42);
}

//+The+function+returning+the+pointer+to+a+function:
Func+*returnAfunc(void)
{
++++return+exampleFunc;
}|code-block|syntax|javascript|1314603|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|$]]

I'm assuming C here (no objects) :) :

<pre><code>// Type of function which takes a char and returns an int:
typedef int (*Func)(char a);

// An example of the function you're trying to return and which does something
// with char:
int exampleFunc(char a)
{
 return (int)(a + 42);
}

// The function returning the pointer to a function:
Func *returnAfunc(void)
{
 return exampleFunc;
}
</code></pre>

blocks|key|1529846|text|像这样的东西|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1529847|#include+<iostream>

typedef+char+(*fn_ptr_t)(char);

char+a_fn(char+c)
{
++return+c+%2B+1;
}

char+b_fn(char+c)
{
++return+c+%2B+2;
}

fn_ptr_t
return_function(char+c)
{
++fn_ptr_t+result+=+0;

++switch+(c)
+{
++++case+'a':
++++++result+=+a_fn;
++++++break;
++++case+'b':
++++++result+=+b_fn;
++++++break;
+}

+return+result;
}

int
main()
{
++fn_ptr_t+fn+=+return_function('a');

++std::cout+<<+"a(l)+=+"+<<+(fn)('l')+<<+std::endl;

++return+0;
}|code-block|syntax|javascript|1529848|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|$]]

Something like this

<pre><code>#include &lt;iostream&gt;

typedef char (*fn_ptr_t)(char);

char a_fn(char c)
{
 return c + 1;
}

char b_fn(char c)
{
 return c + 2;
}

fn_ptr_t
return_function(char c)
{
 fn_ptr_t result = 0;

 switch (c)
 {
 case 'a':
 result = a_fn;
 break;
 case 'b':
 result = b_fn;
 break;
 }

 return result;
}

int
main()
{
 fn_ptr_t fn = return_function('a');

 std::cout &lt;&lt; "a(l) = " &lt;&lt; (fn)('l') &lt;&lt; std::endl;

 return 0;
}
</code></pre>

I'm trying to make a function that takes a character, then returns a pointer to a function depending on what the character was. I just am not sure how to make a function return a pointer to a function.

How to make a function return a pointer to a function? (C++)

我正在尝试创建一个函数，它接受一个字符，然后返回一个指向函数的指针，这取决于该字符是什么。我只是不确定如何让函数返回指向函数的指针。

问如何让函数返回指向函数的指针？(C++)
EN

回答 12

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问如何让函数返回指向函数的指针？(C++)EN

回答 12

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问如何让函数返回指向函数的指针？(C++)
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