std::is_pointer_interconvertible_with_class

template<class S, class M>
constexpr bool is_pointer_interconvertible_with_class( M S::* mp ) noexcept;

Given an object s of type S, determines whether s.*mp refers to a subobject of s and s is pointer-interconvertible with its subobject s.*mp. The program is ill-formed if S is not a complete type.

If S is not a StandardLayoutType, or M is not an object type, or mp is equal to nullptr, the result is always false.

Parameters

Return value

true if s.*mp refers a subobject of s and s is pointer-interconvertible with its subobject s.*mp, otherwise false, where s is an object of type S.

Notes

The type of a pointer-to-member expression &S::m is not always M S::*, where m is of type M, because m may be a member inherited from a base class of S. The template arguments can be specified in order to avoid potentially surprising results.

If there is a value mp of type M S::* such that std::is_pointer_interconvertible_with_class(mp) == true, then reinterpret_cast<M&>(s) has well-defined result and it refers the same subobject as s.*mp, where s is a valid lvalue of type S.

On common platforms, the bit pattern of mp is all zero if std::is_pointer_interconvertible_with_class(mp) == true.

Feature testing macro: __cpp_lib_is_pointer_interconvertible.

Example

#include <type_traits>
#include <iostream>
 
struct Foo { int x; };
struct Bar { int y; };
 
struct Baz : Foo, Bar {}; // not standard-layout
 
int main()
{
    std::cout << std::boolalpha
        << std::is_same_v<decltype(&Baz::x), int Baz::*> << '\n'
        << std::is_pointer_interconvertible_with_class(&Baz::x) << '\n'
        << std::is_pointer_interconvertible_with_class<Baz, int>(&Baz::x) << '\n';
}

false
true
false

See also