20 General utilities library [utilities]

20.15 Metaprogramming and type traits [meta]

20.15.9 Member relationships [meta.member]

đź”—
template<class S, class M> constexpr bool is_pointer_interconvertible_with_class(M S::*m) noexcept;
1
#
Mandates: S is a complete type.
2
#
Returns: true if and only if S is a standard-layout type, M is an object type, m is not null, and each object s of type S is pointer-interconvertible ([basic.compound]) with its subobject s.*m.
đź”—
template<class S1, class S2, class M1, class M2> constexpr bool is_corresponding_member(M1 S1::*m1, M2 S2::*m2) noexcept;
3
#
Mandates: S1 and S2 are complete types.
4
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Returns: true if and only if S1 and S2 are standard-layout types, M1 and M2 are object types, m1 and m2 are not null, and m1 and m2 point to corresponding members of the common initial sequence ([class.mem]) of S1 and S2.
5
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[ Note
:
The type of a pointer-to-member expression &C​::​b is not always a pointer to member of C, leading to potentially surprising results when using these functions in conjunction with inheritance.
[ Example
: 
struct A { int a; };                    // a standard-layout class
struct B { int b; };                    // a standard-layout class
struct C: public A, public B { };       // not a standard-layout class

static_assert( is_pointer_interconvertible_with_class( &C::b ) );
  // Succeeds because, despite its appearance, &C​::​b has type
  // “pointer to member of B of type int”.
static_assert( is_pointer_interconvertible_with_class<C>( &C::b ) );
  // Forces the use of class C, and fails.

static_assert( is_corresponding_member( &C::a, &C::b ) );
  // Succeeds because, despite its appearance, &C​::​a and &C​::​b have types
  // “pointer to member of A of type int” and
  // “pointer to member of B of type int”, respectively.
static_assert( is_corresponding_member<C, C>( &C::a, &C::b ) );
  // Forces the use of class C, and fails.
— end example
 ]
— end note
 ]