获取函数成员地址 [英] get function member address

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;

推荐答案

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)();

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