获取函数成员地址
问题描述：
有没有办法获得函数成员的确切地址？
例如我有：
Is there any way to get the exact address of a function member? For example I have :
struct foo
{
void print() { printf("bla bla bla"); }
};
//////////////////////////////////////////////////////////////
unsigned int address = foo::print;
答
您可以使用以下语法声明指针成员函数：
You can use the following syntax to declare the pointer to the member function:
typedef void (foo::*address)();
address func = &foo::print;
为了调用非静态成员函数，您将需要该类的现有实例：
In order to call non-static member function you will need an existing instance of that class:
(fooInstance.*func)();