blocks|key|1094449|text|语法错误。成员指针是与普通指针不同的类型类别。成员指针必须与其类的对象一起使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1094450|class+A+{
public:
+int+f();
+int+(A::*x)();+//+<-+declare+by+saying+what+class+it+is+a+pointer+to
};

int+A::f()+{
+return+1;
}


int+main()+{
+A+a;
+a.x+=+&A::f;+//+use+the+::+syntax
+printf("%25d\n",(a.*(a.x))());+//+use+together+with+an+object+of+its+class
}|code-block|syntax|javascript|1094451|a.x还没有说明要在哪个对象上调用该函数。它只是说明您想要使用存储在对象a中的指针。另一次将a作为左操作数添加到.*运算符将告诉编译器在哪个对象上调用函数。|offset|length|style|CODE|1094452|entityMap^0|0|0|0|3|10|1|1A|1|1K|2|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]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]|$I|T|J|U|K|L]|$I|V|J|W|K|L]|$I|X|J|Y|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|N|$]]

The syntax is wrong. A member pointer is a different type category from a ordinary pointer. The member pointer will have to be used together with an object of its class:

<pre><code>class A {
public:
 int f();
 int (A::*x)(); // &lt;- declare by saying what class it is a pointer to
};

int A::f() {
 return 1;
}


int main() {
 A a;
 a.x = &amp;A::f; // use the :: syntax
 printf("%d\n",(a.*(a.x))()); // use together with an object of its class
}
</code></pre>

<code>a.x</code> does not yet say on what object the function is to be called on. It just says that you want to use the pointer stored in the object <code>a</code>. Prepending <code>a</code> another time as the left operand to the <code>.*</code> operator will tell the compiler on what object to call the function on.

blocks|key|1094479|text|您需要使用指向成员函数的指针，而不仅仅是指向函数的指针。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1094480|class+A+{+
++++int+f()+{+return+1;+}
public:
++++int+(A::*x)();

++++A()+:+x(&A::f)+{}
};

int+main()+{+
+++A+a;
+++std::cout+<<+(a.*a.x)();
+++return+0;
}|code-block|syntax|javascript|1094481|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|$]]

You need to use a pointer to a member function, not just a pointer to a function.

<pre><code>class A { 
 int f() { return 1; }
public:
 int (A::*x)();

 A() : x(&amp;A::f) {}
};

int main() { 
 A a;
 std::cout &lt;&lt; (a.*a.x)();
 return 0;
}
</code></pre>

blocks|key|1309713|text|int+(*x)()不是指向成员函数的指针。指向成员函数的指针编写如下：int+(A::*x)(void)+=+&A::f;。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1309714|entityMap^0|0|A|10|Q|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|M|8|@]|D|@]|E|$]]]|G|$]]

<code>int (*x)()</code> is not a pointer to member function. A pointer to member function is written like this: <code>int (A::*x)(void) = &amp;A::f;</code>.

blocks|key|1094540|text|Call+member+function+on+string+command|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1094541|#include+<iostream>
#include+<string>


class+A+
{
public:+
++++void+call();
private:
++++void+printH();
++++void+command(std::string+a,+std::string+b,+void+(A::*func)());
};

void+A::printH()
{
++++std::cout<<+"H\n";
}

void+A::call()
{
++++command("a","a",+&A::printH);
}

void+A::command(std::string+a,+std::string+b,+void+(A::*func)())
{
++++if(a+==+b)
++++{
++++++++(this->*func)();
++++}
}

int+main()
{
++++A+a;
++++a.call();
++++return+0;
}|code-block|syntax|javascript|1094542|注意(this->*func)();和用类名void+(A::*func)()声明函数指针的方法|style|CODE|1094543|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/37499889/call-member-function-on-string-command^0|0|12|0|0|0|2|G|M|H|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|Z|8|@$A|10|B|11|K|L]|$A|12|B|13|K|L]]|9|@]|C|$]]|$1|M|3|-4|5|6|7|14|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

<a href="https://stackoverflow.com/questions/37499889/call-member-function-on-string-command">Call member function on string command</a>

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


class A 
{
public: 
 void call();
private:
 void printH();
 void command(std::string a, std::string b, void (A::*func)());
};

void A::printH()
{
 std::cout&lt;&lt; "H\n";
}

void A::call()
{
 command("a","a", &amp;A::printH);
}

void A::command(std::string a, std::string b, void (A::*func)())
{
 if(a == b)
 {
 (this-&gt;*func)();
 }
}

int main()
{
 A a;
 a.call();
 return 0;
}
</code></pre>

Pay attention to <code>(this-&gt;*func)();</code> and the way to declare the function pointer with class name <code>void (A::*func)()</code>

blocks|key|1309791|text|虽然这是基于本页其他地方的标准答案，但我有一个用例并没有被他们完全解决；对于指向函数的指针向量，请执行以下操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1309792|#include+<iostream>
#include+<vector>
#include+<stdio.h>
#include+<stdlib.h>

class+A{
public:
++typedef+vector<int>+(A::*AFunc)(int+I1,int+I2);
++vector<AFunc>+FuncList;
++inline+int+Subtract(int+I1,int+I2){return+I1-I2;};
++inline+int+Add(int+I1,int+I2){return+I1%2BI2;};
++...
++void+Populate();
++void+ExecuteAll();
};

void+A::Populate(){
++++FuncList.push_back(&A::Subtract);
++++FuncList.push_back(&A::Add);
++++...
}

void+A::ExecuteAll(){
++int+In1=1,In2=2,Out=0;
++for(size_t+FuncId=0;FuncId<FuncList.size();FuncId%2B%2B){
++++Out=(this->*FuncList[FuncId])(In1,In2);
++++printf("Function+%25ld+output+%25d\n",FuncId,Out);
++}
}

int+main(){
++A+Demo;
++Demo.Populate();
++Demo.ExecuteAll();
++return+0;
}|code-block|syntax|javascript|1309793|如果你正在编写一个带有索引函数的命令解释器，这些函数需要与参数、语法和帮助提示等结合在一起，这样的东西很有用。可能在菜单中也很有用。|1309794|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|$]]

While this is based on the sterling answers elsewhere on this page, I had a use case which wasn't completely solved by them; for a vector of pointers to functions do the following:

<pre><code>#include &lt;iostream&gt;
#include &lt;vector&gt;
#include &lt;stdio.h&gt;
#include &lt;stdlib.h&gt;

class A{
public:
 typedef vector&lt;int&gt; (A::*AFunc)(int I1,int I2);
 vector&lt;AFunc&gt; FuncList;
 inline int Subtract(int I1,int I2){return I1-I2;};
 inline int Add(int I1,int I2){return I1+I2;};
 ...
 void Populate();
 void ExecuteAll();
};

void A::Populate(){
 FuncList.push_back(&amp;A::Subtract);
 FuncList.push_back(&amp;A::Add);
 ...
}

void A::ExecuteAll(){
 int In1=1,In2=2,Out=0;
 for(size_t FuncId=0;FuncId&lt;FuncList.size();FuncId++){
 Out=(this-&gt;*FuncList[FuncId])(In1,In2);
 printf("Function %ld output %d\n",FuncId,Out);
 }
}

int main(){
 A Demo;
 Demo.Populate();
 Demo.ExecuteAll();
 return 0;
}
</code></pre>

Something like this is useful if you are writing a command interpreter with indexed functions that need to be married up with parameter syntax and help tips etc. Possibly also useful in menus.

blocks|key|1094623|text|不幸的是，您不能将现有的成员函数指针转换为普通函数指针，您可以用一种相当简单的方式创建一个适配器函数模板，该模板将编译时已知的成员函数指针包装在一个普通函数中，如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1094624|template+<class+Type>
struct+member_function;

template+<class+Type,+class+Ret,+class...+Args>
struct+member_function<Ret(Type::*)(Args...)>
{
++++template+<Ret(Type::*Func)(Args...)>
++++static+Ret+adapter(Type+&obj,+Args&&...+args)
++++{
++++++++return+(obj.*Func)(std::forward<Args>(args)...);
++++}
};

template+<class+Type,+class+Ret,+class...+Args>
struct+member_function<Ret(Type::*)(Args...)+const>
{
++++template+<Ret(Type::*Func)(Args...)+const>
++++static+Ret+adapter(const+Type+&obj,+Args&&...+args)
++++{
++++++++return+(obj.*Func)(std::forward<Args>(args)...);
++++}
};|code-block|syntax|javascript|1094625|int+(*func)(A&)+=+&member_function<decltype(&A::f)>::adapter<&A::f>;|1094626|请注意，为了调用成员函数，必须提供A的实例。|offset|length|style|CODE|1094627|entityMap^0|0|0|0|H|1|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]]|$1|G|3|H|5|D|7|S|8|@]|9|@]|A|$E|F]]|$1|I|3|J|5|6|7|T|8|@$K|U|L|V|M|N]]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

While you unfortunately cannot convert an existing member function pointer to a plain function pointer, you can create an adapter function template in a fairly straightforward way that wraps a member function pointer known at compile-time in a normal function like this:

<pre><code>template &lt;class Type&gt;
struct member_function;

template &lt;class Type, class Ret, class... Args&gt;
struct member_function&lt;Ret(Type::*)(Args...)&gt;
{
 template &lt;Ret(Type::*Func)(Args...)&gt;
 static Ret adapter(Type &amp;obj, Args&amp;&amp;... args)
 {
 return (obj.*Func)(std::forward&lt;Args&gt;(args)...);
 }
};

template &lt;class Type, class Ret, class... Args&gt;
struct member_function&lt;Ret(Type::*)(Args...) const&gt;
{
 template &lt;Ret(Type::*Func)(Args...) const&gt;
 static Ret adapter(const Type &amp;obj, Args&amp;&amp;... args)
 {
 return (obj.*Func)(std::forward&lt;Args&gt;(args)...);
 }
};
</code></pre>

 

<pre><code>int (*func)(A&amp;) = &amp;member_function&lt;decltype(&amp;A::f)&gt;::adapter&lt;&amp;A::f&gt;;
</code></pre>

Note that in order to call the member function, an instance of <code>A</code> must be provided.

blocks|key|1309899|text|@Johannes+Schaub+-+litb有正确的解决方案，但我认为发布一个使用指向成员函数的指针的通用示例也会很有好处。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1309900|std::string+myString{+"Hello+World!"+};
auto+memberFunctionPointer{+&std::string::length+};
auto+myStringLength{+(myString.*memberFunctionPointer)()+};|code-block|syntax|javascript|1309901|C%2B%2B17有一个模板函数，用于调用指向成员函数的指针，如下所示。|1309902|std::invoke(memberFunctionPointer,+myString);|1309903|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/2402607/11039217^0|0|N|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|C|$]]|$1|K|3|L|5|F|7|10|8|@]|9|@]|C|$G|H]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

<a href="https://stackoverflow.com/a/2402607/11039217">@Johannes Schaub - litb</a> has the correct solution, but I thought it would be beneficial to post a generic example of using a pointer to a member function as well.
<pre><code>std::string myString{ &quot;Hello World!&quot; };
auto memberFunctionPointer{ &amp;std::string::length };
auto myStringLength{ (myString.*memberFunctionPointer)() };
</code></pre>
C++17 has a template function for calling a pointer to a member function, which looks like this.
<pre><code>std::invoke(memberFunctionPointer, myString);
</code></pre>

I'd like to set up a function pointer as a member of a class that is a pointer to another function in the same class. The reasons why I'm doing this are complicated.

In this example, I would like the output to be "1"

<pre><code>class A {
public:
 int f();
 int (*x)();
}

int A::f() {
 return 1;
}


int main() {
 A a;
 a.x = a.f;
 printf("%d\n",a.x())
}
</code></pre>

But this fails at compiling. Why?

Function pointer to member function

编译器

解释器

我想设置一个函数指针作为类的成员，它是指向同一类中另一个函数的指针。我这么做的原因很复杂。在本例中，我希望输出为"1“class A {public: int f(); int (*x)();}int A::f() { return 1;}int main() { A a; a.x = a.f; printf("%d\...

问指向成员函数的函数指针
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