問題描述

看起來std::cout不能打印成員函數(shù)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
", &TestClass::MyFunc);
  cout << &TestClass::MyFunc << endl;
}

int main(void)
{
  TestClass a;

  a.PrintMyFuncAddress();

  return EXIT_SUCCESS;
}

結果是這樣的:

003111DB
1

如何使用 std::cout 打印 MyFunc 的地址?

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

推薦答案

我不認為該語言提供了任何用于執(zhí)行此操作的工具.operator << 有用于流打印出普通 void* 指針的重載，但成員函數(shù)指針不能轉換為 void*s.這都是特定于實現(xiàn)的，但通常成員函數(shù)指針被實現(xiàn)為一對值 - 一個指示成員函數(shù)是否為虛擬的標志，以及一些額外的數(shù)據(jù).如果函數(shù)是非虛擬函數(shù)，則額外信息通常是實際成員函數(shù)的地址.如果該函數(shù)是虛函數(shù)，則該額外信息可能包含有關如何索引到虛函數(shù)表中以查找給定接收者對象的函數(shù)的數(shù)據(jù).

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.

總的來說，我認為這意味著在不調用未定義行為的情況下打印成員函數(shù)的地址是不可能的.您可能必須使用一些特定于編譯器的技巧才能實現(xiàn)此效果.

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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