std::addressof - strange implementation

Question

So a few days ago i learned about std::addressof. At http://en.cppreference.com/w/cpp/memory/addressof a possible implementation is given:

template< class T >
T* addressof(T& arg) 
{
    return reinterpret_cast<T*>(
               &const_cast<char&>(
                  reinterpret_cast<const volatile char&>(arg)));
}

As far i see this can simply be implemented like:

template<typename T>
T* addressof( T& var )
{
    return &var;
}

Why the guys at cppreference chose to implement it with 3 casts? Is there any detail I am missing that is making their implementation better. What is the point in using volatile when all you do is cast?

Overloading the address of operator breaks the second implementation. — SirGuy, Jul 28 '15 at 17:15
have a look at this as well: http://stackoverflow.com/questions/6494591/how-can-i-reliably-get-the-address-of-an-object/6495205#6495205 — NathanOliver, Jul 28 '15 at 17:16

score 9 · Accepted Answer · answered Jul 28 '15 at 17:15

9

If it could be implemented like in your example, there would be no need for it. The point is that it gives you the address of an object, even if the address-of operator& for that type has been overloaded.

answered Jul 28 '15 at 17:15

juanchopanza

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Aha I see. And what is the point of casting to const volatile&, instead of just char&? – GeneralFailure Jul 28 '15 at 17:19
@GeneralFailure That is so that you can pass `const` or `volatile` qualified lvalues. You can try removing that to see. – juanchopanza Jul 28 '15 at 17:35

std::addressof - strange implementation

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