9 Declarations [dcl]

9.2 Specifiers [dcl.spec]

9.2.9 Type specifiers [dcl.type]

9.2.9.6 Decltype specifiers [dcl.type.decltype]

1
#
For an expression E, the type denoted by decltype(E) is defined as follows:
  • (1.1)
    if E is an unparenthesized id-expression naming a structured binding ([dcl.struct.bind]), decltype(E) is the referenced type as given in the specification of the structured binding declaration;
  • (1.2)
    otherwise, if E is an unparenthesized id-expression naming a constant template parameter ([temp.param]), decltype(E) is the type of the template parameter after performing any necessary type deduction ([dcl.spec.auto], [dcl.type.class.deduct]);
  • (1.3)
    otherwise, if E is an unparenthesized id-expression or an unparenthesized class member access ([expr.ref]), decltype(E) is the type of the entity named by E.
    If there is no such entity, the program is ill-formed;
  • (1.4)
    otherwise, if E is an unparenthesized splice-expression, decltype(E) is the type of the entity, object, or value designated by the splice-specifier of E;
  • (1.5)
    otherwise, if E is an xvalue, decltype(E) is T&&, where T is the type of E;
  • (1.6)
    otherwise, if E is an lvalue, decltype(E) is T&, where T is the type of E;
  • (1.7)
    otherwise, decltype(E) is the type of E.
The operand of the decltype specifier is an unevaluated operand.
[Example 1: const int&& foo(); int i; struct A { double x; }; const A* a = new A(); decltype(foo()) x1 = 17; // type is const int&& decltype(i) x2; // type is int decltype(a->x) x3; // type is double decltype((a->x)) x4 = x3; // type is const double& decltype([:^^x1:]) x5 = 18; // type is const int&& decltype(([:^^x1:])) x6 = 19; // type is const int& void f() { [](auto ...pack) { decltype(pack...[0]) x7; // type is int decltype((pack...[0])) x8; // type is int& }(0); } — end example]
[Note 1: 
The rules for determining types involving decltype(auto) are specified in [dcl.spec.auto].
— end note]
2
#
If the operand of a decltype-specifier is a prvalue and is not a (possibly parenthesized) immediate invocation ([expr.const]), the temporary materialization conversion is not applied ([conv.rval]) and no result object is provided for the prvalue.
The type of the prvalue may be incomplete or an abstract class type.
[Note 2: 
As a result, storage is not allocated for the prvalue and it is not destroyed.
Thus, a class type is not instantiated as a result of being the type of a function call in this context.
In this context, the common purpose of writing the expression is merely to refer to its type.
In that sense, a decltype-specifier is analogous to a use of a typedef-name, so the usual reasons for requiring a complete type do not apply.
In particular, it is not necessary to allocate storage for a temporary object or to enforce the semantic constraints associated with invoking the type's destructor.
— end note]
[Note 3: 
Unlike the preceding rule, parentheses have no special meaning in this context.
— end note]
[Example 2: template<class T> struct A { ~A() = delete; }; template<class T> auto h() -> A<T>; template<class T> auto i(T) // identity -> T; template<class T> auto f(T) // #1 -> decltype(i(h<T>())); // forces completion of A<T> and implicitly uses A<T>​::​~A() // for the temporary introduced by the use of h(). // (A temporary is not introduced as a result of the use of i().) template<class T> auto f(T) // #2 -> void; auto g() -> void { f(42); // OK, calls #2. (#1 is not a viable candidate: type deduction // fails ([temp.deduct]) because A<int>​::​~A() is implicitly used in its // decltype-specifier) } template<class T> auto q(T) -> decltype((h<T>())); // does not force completion of A<T>; A<T>​::​~A() is not implicitly // used within the context of this decltype-specifier void r() { q(42); // error: deduction against q succeeds, so overload resolution selects // the specialization “q(T) -> decltype((h<T>()))'' with Tint; // the return type is A<int>, so a temporary is introduced and its // destructor is used, so the program is ill-formed } — end example]