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std::is_pointer_interconvertible_base_of(3) C++ Standard Libary 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)

2024.06.10 http://cppreference.com