为什么不能用它的名字打印成员函数的地址？

Question

I have learned that function name equals function address like this:

void func(){}
void main() { cout << func; }

But when I used the same code to print memeber function, it went wrong.

class Test{
public:
    void func() {}
    void printFunc1() {
    cout << func << endl;
   }
    void printFunc2() {
    void (Test::*ptrtofn)() = &Test::func;
    cout << (void*&)ptrtofn << endl;
  }
};

printFunction2() work but printFunction1() doesnt

What makes the difference?

Member function's name is not member function's address? Is there any reason?

Answer 1

member function != standalone function

Only standalone functions can be converted to pointer implicitely.

4.3 Function-to-pointer conversion [conv.func]
1 An lvalue of function type T can be converted to a prvalue of type “pointer to T.” The result is a pointer to the function. 58

58) This conversion never applies to non-static member functions because an lvalue that refers to a non-static member function cannot be obtained.

Answer 2

Please understand "func" is the member function of the class . accessing it directly is itself a compilation error .Rather you should try to use pointer to member function as you have done in printFunction2: Else if func is function outside the class scope .Then it can be done as below :

#include <iostream>
using namespace std;
 void func() {cout<<"\n calling func\n";}
    void printFunc1() {
       cout << endl<<hex<<(void*)func << endl;
    }
int main() {

    printFunc1();
    return 0;
}