std::is_pointer_interconvertible_base_of

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Defined in header <type_traits>
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

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

value
[static]
 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

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

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

(C++11)
 checks if a type is a base of the other type
(class template)
(C++11)
 checks if a type is a class (but not union) type and has no non-static data members
(class template)
(C++11)
 checks if a type is a standard-layout type
(class template)
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