Is printing a member pointer a user-defined

Question

Suppose I have this code:

#include <iostream> struct Mine { int a; int b; }; int main() { int Mine::* memberPointerA = &Mine::a; int Mine::* memberPointerB = &Mine::b; std::cout << memberPointerA; std::cout << "\n"; std::cout << memberPointerB; } 

When I run this using Microsoft Visual C ++ (2015)

I get the following output

 1 1 

The result I expect looks something like this:

 1 2 

So this asks the question: is this printing the element pointer in a certain way?

Answer 1

There is defined conversion from pointer to bool . Since element variable pointers are not NULL , they are evaluated as true and printed as 1 .

Answer 2

The key problem is that a member pointer cannot be converted to void* , which is what overloading does, which usually processes print pointers. A.

Thus, the following best transform is used, which is a pointer to a transform → bool . Both pointers are not null pointers, so you get the output you see.

If you try to print “regular” pointers (as opposed to pointers to an element), you will get some output along the lines of what you originally expected.