From Jason Turner

[dcl.type.decltype]

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