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;
}
the result is something like this:
003111DB
1
How can I print MyFunc
's address using std::cout
?
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.
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.
Hope this helps!