std::is_pointer_interconvertible_base_of(3)

NAME
std::is_pointer_interconvertible_base_of - std::is_pointer_interconvertible_base_of

Synopsis
   Defined in header <type_traits>
   template< class Base, class Derived > (since C++20)
   struct is_pointer_interconvertible_base_of;
   If Derived is unambiguously derived from Base and every Derived object is
   pointer-interconvertible with its Base subobject, or if both are the same non-union
   class (in both cases ignoring cv-qualification), provides the member constant value
   equal to true. Otherwise value is false.
   If both Base and Derived are non-union class types, and they are not the same type
   (ignoring cv-qualification), Derived shall be a complete type; otherwise the
   behavior is undefined.
   If the program adds specializations for std::is_pointer_interconvertible_base_of or
   std::is_pointer_interconvertible_base_of_v, the behavior is undefined.

Member constants
            true if Derived is unambiguously derived from Base and every Derived object
   value is pointer-interconvertible with its Base subobject, or if both are the
   [static] same non-union class (in both cases ignoring cv-qualification), false
            otherwise
            (public static member constant)

Member functions
   operator bool converts the object to bool, returns value
                 (public member function)
   operator() returns value
   (C++14) (public member function)

Member types
   Type Definition
   value_type bool
   type std::integral_constant<bool, value>

Notes
   std::is_pointer_interconvertible_base_of_v<T, U> may be true even if T is a private
   or protected base class of U.
   Let
     * U be a complete object type,
     * T be a complete object type with cv-qualification not less than U,
     * u be any valid lvalue of U,
   reinterpret_cast<T&>(u) always has well-defined result if
   std::is_pointer_interconvertible_base_of_v<T, U> is true.
   If T and U are not the same type (ignoring cv-qualification) and T is a
   pointer-interconvertible base class of U, then both std::is_standard_layout_v<T> and
   std::is_standard_layout_v<U> are true.
   If T is standard layout class type, then all base classes of T (if any) are
   pointer-interconvertible base class of T.
         Feature-test macro Value Std Feature
                        Pointer-interconvertibility traits: __cpp_lib_is_pointer_interconvertible 201907L (C++20) * std::is_pointer_interconvertible_base_of,
                        * std::is_pointer_interconvertible_with_class

Example
// Run this code
 #include <type_traits>
 struct Foo {};
 struct Bar {};
 class Baz : Foo, public Bar { int x; };
 class NonStdLayout : public Baz { int y; };
 static_assert(std::is_pointer_interconvertible_base_of_v<Bar, Baz>);
 static_assert(std::is_pointer_interconvertible_base_of_v<Foo, Baz>);
 static_assert(not std::is_pointer_interconvertible_base_of_v<Baz, NonStdLayout>);
 static_assert(std::is_pointer_interconvertible_base_of_v<NonStdLayout, NonStdLayout>);
 int main() {}

See also
   is_base_of checks if a type is derived from the other type
   (C++11) (class template)
   is_empty checks if a type is a class (but not union) type and has no
   (C++11) non-static data members
                      (class template)
   is_standard_layout checks if a type is a standard-layout type
   (C++11) (class template)