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.
希望这有助于！
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