函数指针

（1/2）类型

普通函数指针不能被赋值为成员函数的地址。

int (*pFunc)();

pFunc是一个函数指针，而int (*)()是类型。

成员函数地址要赋值给成员函数指针。

class Base
{
public:
int func() { return 1; }
};

int main()
{
int (*pFunc)();
pFunc = &Base::func;  //error: cannot convert 'int (Base::*)()' to 'int (*)()' in assignment
return 0;
}

class Base
{
public:
int func() { return 1; }
};

int main()
{
int (Base::*pFunc)(); // OK
pFunc = &Base::func;
return 0;
}

再来：

class Base
{
public:
int func() { return 1; }
};
class Derived : public Base
{
public:
int foo() { return 1; }
};

int main()
{
typedef int (Base::*FuncPtr)();
FuncPtr fPtr = &Derived::foo; // error
return 0;
}

需进行强制类型转换

......

FuncPtr fPtr = (FuncPtr)&Derived::foo;  // OK

反过来，基类成员函数地址可赋值给指向派生类成员函数的指针。

class Base
{
public:
int func() { return 1; }
};
class Derived : public Base
{
public:
int foo() { return 1; }
};

int main()
{
typedef int (Derived::*FuncPtr)();
FuncPtr fPtr = &Base::func;  // OK
return 0;
}

（2/2）调用

class Base
{
public:
virtual void func(int i) { cout << "Base: " << i << endl; }
};
class Derived : public Base
{
public:
virtual void func(int i) { cout << "Derived: " << i << endl; }
};

int main()
{
typedef void (Base::*FuncPtr)(int);
FuncPtr fPtr = &Base::func;
Base b;
(b.*fPtr)(2);
Derived* dPtr = new Derived();
(dPtr->*fPtr)(2);
Base* bPtr = new Derived();
(bPtr->*fPtr)(2);
return 0;
}

output:

Base: 2
Derived: 2
Derived: 2