函数指针获取成员函数地址（包括虚函数）

++

class Base
{
public:
	void func() { printf("Enter Base::func()\n"); }
	virtual void virtual_func() { printf("Enter Base::virtual_func()\n"); }
};

class Derived : public Base
{
public:
	void func() { printf("Enter Derived::func()\n"); }
	virtual void virtual_func() { printf("Enter Derived::virtual_func()\n"); }
};

typedef void (Base::*p_base_func)();

int main(int argc, char* argv[]) 
{ 
	Derived d;
	Base b;
	Base *pb = &d;

	p_base_func pbf = &Base::func;
	printf("-- Call via pointer to normal function of base\n");
	(pb->*pbf)();
	(d.*pbf)();
	cout << endl;
	//cout << (pb->*pbf) << endl;

	cout << endl << endl;
	pbf = &Base::virtual_func;
	printf("-- Call via pointer to virtual function of base\n");
	(pb->*pbf)();
	(d.*pbf)();
	//cout << (pb->*pbf) << endl;

	system("pause");
	return 0; 
}

注释部分会发生编译器内部错误。我也不知道为什么？

看代码，为成员函数添加一个引用（支持虚函数）。
然后利用对象调用函数。
貌似，只能利用对象才能成功。