Why does printing address of data member of a struct enabled, although there is no object?

Question

Suppose I have the following code. I have expected that this has to at least give me a warning.

#include <iostream>

struct test
{
    int whatever;
};

int main()
{
    int test::* p = &test::whatever;    
    std::cout << p;
}

However, I was surprised to find out that this code compiles without complaints. So I would like to know, what is actually printed in this case, how is the test::whatever stored, so we can access its address?

Does this answer your question? [Pointer to class data member "::\*"](https://stackoverflow.com/questions/670734/pointer-to-class-data-member) — Botje, Aug 31 '21 at 11:21
OP asks why it is possible to make a pointer of something that does not exist — Raildex, Aug 31 '21 at 11:22

score 3 · Accepted Answer · answered Aug 31 '21 at 11:22

how is the test::whatever stored

This is not specified in the language. But we can make a reasonable guess that it stores an offset from the beginning of the object to the pointed member.

so we can access its address?

No.

what is actually printed in this case

1 is printed.

There is no operator overload for member pointers. But there is an overload for bool and member pointers are convertible to bool. They convert to true when they aren't null. p isn't null, so it converts to true which prints as 1.

Özgür Murat Sağdıçoğlu · Answer 2 · 2021-08-31T11:43:05.263

1

Here p is pointer to member. It is like an offset of the class member relative to the object base address. It doesn't point to anywhere in memory by itself. It needs a base object pointer to point somewhere. Therefore it is not illegal to have a pointer to member without an object, just the opposite: it is the use case of this type.

You can use it like:

#include <iostream>

struct test
{
    int whatever;
};

int main()
{
    int test::* p = &test::whatever;

    test t = {5};
    std::cout << t.*p << std::endl; // prints 5
}

However, I don't know what is printed out in the example you gave.

See Pointer-to-Member ->* and .* Operators in C++.

edited Aug 31 '21 at 11:43

answered Aug 31 '21 at 11:21

Özgür Murat Sağdıçoğlu

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I know what it does. I wanted to know why the code above compiles. – Karen Baghdasaryan Aug 31 '21 at 11:32
Because it is the use case of this type of pointer. You store and pass the relative address of a member to base address of an object of the class/struct. And it is only meaningful when you have a real object. @KarenBaghdasaryan – Özgür Murat Sağdıçoğlu Aug 31 '21 at 11:51

armagedescu · Answer 3 · 2021-08-31T14:47:36.683

It is member pointer, as already pointed in first answer. You can use it to point to different members of different objects:


struct test
{
    int whatever;
};

int main()
{
    int test::* p = &test::whatever;
    test w1{ 123 }, w2{ 234 };
    std::cout << p<< endl; //point to member
    std::cout << w1.*p << endl; //value of first whatever
    std::cout << w2.*p << endl; //value of second whatever

    //real address of whatever of w1 and w2, compare
    std::cout << &(w1.*p) << "=="<< &w1.whatever<< endl;
    std::cout << &(w2.*p) << "=="<< &w2.whatever<< endl;

    std::cout << p << endl; //yet pointer did not change
    return 0;
}

Why does printing address of data member of a struct enabled, although there is no object?

3 Answers3