如何在C ++中打印成员函数地址 [英] How to print member function address in C++

问题描述

看起来 std :: cout 无法打印成员函数的地址，例如：

It looks like std::cout can't print member function's address, for example:

#include <iostream>

using std::cout;
using std::endl;

class TestClass
{
  void MyFunc(void);

public:
  void PrintMyFuncAddress(void);
};

void TestClass::MyFunc(void)
{
  return;
}

void TestClass::PrintMyFuncAddress(void)
{
  printf("%p\n", &TestClass::MyFunc);
  cout << &TestClass::MyFunc << endl;
}

int main(void)
{
  TestClass a;

  a.PrintMyFuncAddress();

  return EXIT_SUCCESS;
}

结果如下：

003111DB
1

我如何使用 std :: cout 打印 MyFunc 的地址？

How can I print MyFunc's address using std::cout?

推荐答案

我不相信有这样的语言提供任何设施。对于流打印出正常的 void * 指针，有 operator 的重载，但成员函数指针不能转换为 void * 。这是所有实现特定的，但是通常成员函数指针被实现为一对值 - 指示成员函数是否是虚拟的标志和一些额外的数据。如果函数是非虚函数，那么额外信息通常是实际成员函数的地址。如果函数是虚函数，那么额外的信息可能包含关于如何索引到虚函数表中的数据，以找到给定接收者对象调用的函数。

I don't believe that there are any facilities provided by the language for doing this. There are overloads for operator << for streams to print out normal void* pointers, but member function pointers are not convertible to void*s. This is all implementation-specific, but typically member function pointers are implemented as a pair of values - a flag indicating whether or not the member function is virtual, and some extra data. If the function is a non-virtual function, that extra information is typically the actual member function's address. If the function is a virtual function, that extra information probably contains data about how to index into the virtual function table to find the function to call given the receiver object.

一般，我认为这意味着不可能打印出成员函数的地址，而不调用未定义的行为。你可能必须使用一些编译器特定的技巧来实现这种效果。

In general, I think this means that it's impossible to print out the addresses of member functions without invoking undefined behavior. You'd probably have to use some compiler-specific trick to achieve this effect.

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