std::is_pointer_interconvertible_base_of

std::is_pointer_interconvertible_base_of - cppreference.com

From cppreference.com


`template< class Base, class Derived > struct is_pointer_interconvertible_base_of;`		(since C++20)

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.

Helper variable template

`template< class Base, class Derived > inline constexpr bool is_pointer_interconvertible_base_of_v = is_pointer_interconvertible_base_of<Base, Derived>::value;`		(since C++20)

Inherited from std::integral_constant

Member constants

true 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), false otherwise
(public static member constant)

Member functions

	converts the object to `bool`, returns `value` (public member function)
	returns `value` (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
`__cpp_lib_is_pointer_interconvertible`	`201907L`	(C++20)	Pointer-interconvertibility traits: `std::is_pointer_interconvertible_base_of`, std::is_pointer_interconvertible_with_class

Example

#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() {}