如何将一个类中（或其对象）的函数的入口地址输出？？
一下为本人测试的一个程序，通不过编译！！
#include <iostream.h>
class A
{
public:
void (*p)();
A(){p=f1;} //出错！！！！
private:
void f1() {cout<<"a^^^^^^^"<<endl;}
};
void main()
{
A a;
cout<<a.p<<" A::f1"<<endl;

}
错误：
error C2440: '=' : cannot convert from 'void (__thiscall A::*)(void)' to 'void (__cdecl *)(void)'
There is no context in which this conversion is possible
执行 cl.exe 时出错.

[此贴子已经被作者于2007-4-11 12:21:52编辑过]

我直接将 void (*p)();换成void (__thiscall A::*p)();，结果如下了：
E:\SourceFiles\testFunction.cpp(5) : error C4234: nonstandard extension used : '__thiscall' keyword reserved for future use

其实我做这个程序的目的在于观察对象中各个成员在内存中的组织形式，也就是想看看类的不同对象之间有哪些是关联在一起的，或者说想搞清是通过地址调用共享一部分成员，还是给每个对象都开辟内存生成变量和函数。

This is the default calling convention used by C++ member functions that do not use variable arguments. The callee cleans the stack, so the compiler makes vararg functions __cdecl, and pushes the this pointer on the stack last. The thiscall calling convention cannot be explicitly specified in a program, because thiscall is not a keyword.