std::is_pointer_interconvertible_base_of

Since C++20

Header: <type_traits>

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.

# Declarations

template< class Base, class Derived >
struct is_pointer_interconvertible_base_of;

(since C++20)

# 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

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 and std::is_standard_layout_v are true.

If T is standard layout class type, then all base classes of T (if any) are pointer-interconvertible base class of T.

# 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() {}

# See also