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.
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.
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.
#include <type_traits>
struct Foo { int x; };
struct Bar { int y; };
struct Baz : Foo, Bar {}; // not standard-layout
static_assert( not std::is_same_v<decltype(&Baz::x), int Baz::*> );
static_assert( std::is_pointer_interconvertible_with_class(&Baz::x) );
static_assert( not std::is_pointer_interconvertible_with_class<Baz, int>(&Baz::x) );
int main() { }[edit] See also
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