如何定义指向静态成员函数的函数指针？

Question

#include "stdafx.h"

class Person;
typedef void (Person::*PPMF)();

// error C2159: more than one storage class specified
typedef static void (Person::*PPMF2)();  

class Person
{
public:
    static PPMF verificationFUnction()
    { 
        return &Person::verifyAddress; 
    }

    // error C2440: 'return' : cannot convert from 
    // 'void (__cdecl *)(void)' to 'PPMF2'
    PPMF2 verificationFUnction2()               
    { 
        return &Person::verifyAddress2; 
    }
private:
    void verifyAddress() {}

    static void verifyAddress2() {}
};

int _tmain(int argc, _TCHAR* argv[])
{
    Person scott;

    PPMF pmf = scott.verificationFUnction();
    (scott.*pmf)();
    return 0;
}

问题:我需要定义一个函数指针PPMF2来指向一个静态成员函数verifyAddress2.我该怎么做？

#include "stdafx.h"

class Person;
typedef void (Person::*PPMF)();
typedef void (Person::*PPMF2)();

class Person
{
public:
    static PPMF verificationFUnction()
    { 
        return &Person::verifyAddress; 
    }
    PPMF2 verificationFUnction2()
    { 
        return &Person::verifyAddress2; 
    }
private:
    void verifyAddress() {}

    static void verifyAddress2() {}
};

int _tmain(int argc, _TCHAR* argv[])
{
    Person scott;

    PPMF pmf = scott.verificationFUnction();
    (scott.*pmf)();

    return 0;
}

Answer 1

指向静态成员函数的指针只是一个普通的函数指针.typedef void (*PPMF2)().您可以将它分配给静态成员函数,就像分配任何函数指针一样,只是静态成员函数在类范围内:

PPMF2 myfunc = &MyClass::StaticMemberFunc;

Answer 2

关于静态成员函数保证:

С++ ISO/IEC 14882 2003-10-15说

5.2.2函数调用有两种:普通函数调用和成员函数57)(9.3)调用....

57)静态成员函数(9.4)是普通函数.

理论上,static-member-functions可以有另一个调用约定.但标准允许我们利用这样的事情......

答案:typedef void(Person ::*PPMF2)()=> typedef void(*PPMF2)()