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.

The behavior of a program that adds specializations for
is_pointer_interconvertible_base_of or is_pointer_interconvertible_base_of_v is
undefined.

Helper variable template

template< class Base, class Derived >

inline constexpr bool is_pointer_interconvertible_base_of_v = (since C++20)

is_pointer_interconvertible_base_of<Base, Derived>::value;

Inherited from std::integral_constant

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: __cpp_lib_is_pointer_interconvertible

Example

// Run this code

#include <iostream>
#include <type_traits>

struct Foo {};

struct Bar {};

class Baz : Foo, public Bar {
int x;
};

class NonStdLayout : public Baz {
int y;
};

int main()
{
std::cout << std::boolalpha
<< std::is_pointer_interconvertible_base_of_v<Bar, Baz> << '\n'
<< std::is_pointer_interconvertible_base_of_v<Foo, Baz> << '\n'
<< std::is_pointer_interconvertible_base_of_v<Baz, NonStdLayout> << '\n'
<< std::is_pointer_interconvertible_base_of_v<NonStdLayout, NonStdLayout> << '\n';
}

Output:

true
true
false
true

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)