7 Declarations [dcl.dcl]

There are two cv-qualifiers, const and volatile. Each cv-qualifier shall appear at most once in a cv-qualifier-seq. If a cv-qualifier appears in a decl-specifier-seq, the init-declarator-list of the declaration shall not be empty. [ Note: [basic.type.qualifier] and [dcl.fct] describe how cv-qualifiers affect object and function types.  — end note ] Redundant cv-qualifications are ignored. [ Note: For example, these could be introduced by typedefs. — end note ]

[ Note: Declaring a variable const can affect its linkage ([dcl.stc]) and its usability in constant expressions ([expr.const]). As described in [dcl.init], the definition of an object or subobject of const-qualified type must specify an initializer or be subject to default-initialization.  — end note ]

A pointer or reference to a cv-qualified type need not actually point or refer to a cv-qualified object, but it is treated as if it does; a const-qualified access path cannot be used to modify an object even if the object referenced is a non-const object and can be modified through some other access path. [ Note: Cv-qualifiers are supported by the type system so that they cannot be subverted without casting ([expr.const.cast]).  — end note ]

Except that any class member declared mutable ([dcl.stc]) can be modified, any attempt to modify a const object during its lifetime ([basic.life]) results in undefined behavior. [ Example:

const int ci = 3;               // cv-qualified (initialized as required)
ci = 4;                         // ill-formed: attempt to modify const

int i = 2;                      // not cv-qualified
const int* cip;                 // pointer to const int
cip = &i;                       // OK: cv-qualified access path to unqualified
*cip = 4;                       // ill-formed: attempt to modify through ptr to const

int* ip;
ip = const_cast<int*>(cip);     // cast needed to convert const int* to int*
*ip = 4;                        // defined: *ip points to i, a non-const object

const int* ciq = new const int (3);     // initialized as required
int* iq = const_cast<int*>(ciq);        // cast required
*iq = 4;                                // undefined: modifies a const object

For another example

struct X {
  mutable int i;
  int j;
};
struct Y {
  X x;
  Y();
};

const Y y;
y.x.i++;                        // well-formed: mutable member can be modified
y.x.j++;                        // ill-formed: const-qualified member modified
Y* p = const_cast<Y*>(&y);      // cast away const-ness of y
p->x.i = 99;                    // well-formed: mutable member can be modified
p->x.j = 99;                    // undefined: modifies a const member

 — end example ]

What constitutes an access to an object that has volatile-qualified type is implementation-defined. If an attempt is made to refer to an object defined with a volatile-qualified type through the use of a glvalue with a non-volatile-qualified type, the program behavior is undefined.

[ Note: volatile is a hint to the implementation to avoid aggressive optimization involving the object because the value of the object might be changed by means undetectable by an implementation. Furthermore, for some implementations, volatile might indicate that special hardware instructions are required to access the object. See [intro.execution] for detailed semantics. In general, the semantics of volatile are intended to be the same in C++ as they are in C.  — end note ]

The simple type specifiers are

simple-type-specifier:
    nested-name-specifieropt type-name
    nested-name-specifier template simple-template-id
    char
    char16_t
    char32_t
    wchar_t
    bool
    short
    int
    long
    signed
    unsigned
    float
    double
    void
    auto
    decltype-specifier

type-name:
    class-name
    enum-name
    typedef-name
    simple-template-id

decltype-specifier:
  decltype ( expression )
  decltype ( auto )

The auto specifier is a placeholder for a type to be deduced ([dcl.spec.auto]). The other simple-type-specifiers specify either a previously-declared type, a type determined from an expression, or one of the fundamental types ([basic.fundamental]). Table [tab:simple.type.specifiers] summarizes the valid combinations of simple-type-specifiers and the types they specify.

When multiple simple-type-specifiers are allowed, they can be freely intermixed with other decl-specifiers in any order. [ Note: It is implementation-defined whether objects of char type are represented as signed or unsigned quantities. The signed specifier forces char objects to be signed; it is redundant in other contexts.  — end note ]

For an expression e, the type denoted by decltype(e) is defined as follows:

• (4.1)

  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, or if e names a set of overloaded functions, the program is ill-formed;

• (4.2)

  otherwise, if e is an xvalue, decltype(e) is T&&, where T is the type of e;

• (4.3)

  otherwise, if e is an lvalue, decltype(e) is T&, where T is the type of e;

• (4.4)

  otherwise, decltype(e) is the type of e.

The operand of the decltype specifier is an unevaluated operand (Clause [expr]).

[ Example:

const int&& foo();
int i;
struct A { double x; };
const A* a = new A();
decltype(foo()) x1 = 0;         // 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&

 — end example ] [ Note: The rules for determining types involving decltype(auto) are specified in [dcl.spec.auto].  — end note ]

[ Note: in the case where the operand of a decltype-specifier is a function call and the return type of the function is a class type, a special rule ([expr.call]) ensures that the return type is not required to be complete (as it would be if the call appeared in a sub-expression or outside of a decltype-specifier). 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. [ Example:

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 T=int]”.
                                // The return type is A<int>, so a temporary is introduced and its
                                // destructor is used, so the program is ill-formed.
}

 — end example ] — end note ]

An attribute-specifier-seq shall not appear in an elaborated-type-specifier unless the latter is the sole constituent of a declaration. If an elaborated-type-specifier is the sole constituent of a declaration, the declaration is ill-formed unless it is an explicit specialization ([temp.expl.spec]), an explicit instantiation ([temp.explicit]) or it has one of the following forms:

class-key attribute-specifier-seqopt identifier ;
friend class-key ::opt identifier ;
friend class-key ::opt simple-template-id ;
friend class-key nested-name-specifier identifier ;
friend class-key nested-name-specifier templateopt simple-template-id ;

In the first case, the attribute-specifier-seq, if any, appertains to the class being declared; the attributes in the attribute-specifier-seq are thereafter considered attributes of the class whenever it is named.

[basic.lookup.elab] describes how name lookup proceeds for the identifier in an elaborated-type-specifier. If the identifier resolves to a class-name or enum-name, the elaborated-type-specifier introduces it into the declaration the same way a simple-type-specifier introduces its type-name. If the identifier resolves to a typedef-name or the simple-template-id resolves to an alias template specialization, the elaborated-type-specifier is ill-formed. [ Note: This implies that, within a class template with a template type-parameter T, the declaration

friend class T;

is ill-formed. However, the similar declaration friend T; is allowed ([class.friend]).  — end note ]

The class-key or enum keyword present in the elaborated-type-specifier shall agree in kind with the declaration to which the name in the elaborated-type-specifier refers. This rule also applies to the form of elaborated-type-specifier that declares a class-name or friend class since it can be construed as referring to the definition of the class. Thus, in any elaborated-type-specifier, the enum keyword shall be used to refer to an enumeration ([dcl.enum]), the union class-key shall be used to refer to a union (Clause [class]), and either the class or struct class-key shall be used to refer to a class (Clause [class]) declared using the class or struct class-key. [ Example:

enum class E { a, b };
enum E x = E::a;                // OK

 — end example ]

The auto and decltype(auto) type-specifiers designate a placeholder type that will be replaced later, either by deduction from an initializer or by explicit specification with a trailing-return-type. The auto type-specifier is also used to signify that a lambda is a generic lambda.

The placeholder type can appear with a function declarator in the decl-specifier-seq, type-specifier-seq, conversion-function-id, or trailing-return-type, in any context where such a declarator is valid. If the function declarator includes a trailing-return-type ([dcl.fct]), that specifies the declared return type of the function. If the declared return type of the function contains a placeholder type, the return type of the function is deduced from return statements in the body of the function, if any.

If the auto type-specifier appears as one of the decl-specifiers in the decl-specifier-seq of a parameter-declaration of a lambda-expression, the lambda is a generic lambda ([expr.prim.lambda]). [ Example:

auto glambda = [](int i, auto a) { return i; }; // OK: a generic lambda

 — end example ]

The type of a variable declared using auto or decltype(auto) is deduced from its initializer. This use is allowed when declaring variables in a block ([stmt.block]), in namespace scope ([basic.scope.namespace]), and in a for-init-statement ([stmt.for]). auto or decltype(auto) shall appear as one of the decl-specifiers in the decl-specifier-seq and the decl-specifier-seq shall be followed by one or more init-declarators, each of which shall have a non-empty initializer. In an initializer of the form

( expression-list )

the expression-list shall be a single assignment-expression.

[ Example:

auto x = 5;                 // OK: x has type int
const auto *v = &x, u = 6;  // OK: v has type const int*, u has type const int
static auto y = 0.0;        // OK: y has type double
auto int r;                 // error: auto is not a storage-class-specifier
auto f() -> int;            // OK: f returns int
auto g() { return 0.0; }    // OK: g returns double
auto h();                   // OK: h's return type will be deduced when it is defined

 — end example ]

A placeholder type can also be used in declaring a variable in the condition of a selection statement ([stmt.select]) or an iteration statement ([stmt.iter]), in the type-specifier-seq in the new-type-id or type-id of a new-expression ([expr.new]), in a for-range-declaration, and in declaring a static data member with a brace-or-equal-initializer that appears within the member-specification of a class definition ([class.static.data]).

A program that uses auto or decltype(auto) in a context not explicitly allowed in this section is ill-formed.

When a variable declared using a placeholder type is initialized, or a return statement occurs in a function declared with a return type that contains a placeholder type, the deduced return type or variable type is determined from the type of its initializer. In the case of a return with no operand, the initializer is considered to be void(). Let T be the declared type of the variable or return type of the function. If the placeholder is the auto type-specifier, the deduced type is determined using the rules for template argument deduction. If the deduction is for a return statement and the initializer is a braced-init-list ([dcl.init.list]), the program is ill-formed. Otherwise, obtain P from T by replacing the occurrences of auto with either a new invented type template parameter U or, if the initializer is a braced-init-list, with std::initializer_list<U>. Deduce a value for U using the rules of template argument deduction from a function call ([temp.deduct.call]), where P is a function template parameter type and the initializer is the corresponding argument. If the deduction fails, the declaration is ill-formed. Otherwise, the type deduced for the variable or return type is obtained by substituting the deduced U into P. [ Example:

auto x1 = { 1, 2 };         // decltype(x1) is std::initializer_list<int>
auto x2 = { 1, 2.0 };       // error: cannot deduce element type

 — end example ]

[ Example:

const auto &i = expr;

The type of i is the deduced type of the parameter u in the call f(expr) of the following invented function template:

template <class U> void f(const U& u);

 — end example ]

If the placeholder is the decltype(auto) type-specifier, the declared type of the variable or return type of the function shall be the placeholder alone. The type deduced for the variable or return type is determined as described in [dcl.type.simple], as though the initializer had been the operand of the decltype. [ Example:

int i;
int&& f();
auto           x3a = i;        // decltype(x3a) is int
decltype(auto) x3d = i;        // decltype(x3d) is int
auto           x4a = (i);      // decltype(x4a) is int
decltype(auto) x4d = (i);      // decltype(x4d) is int&
auto           x5a = f();      // decltype(x5a) is int
decltype(auto) x5d = f();      // decltype(x5d) is int&&
auto           x6a = { 1, 2 }; // decltype(x6a) is std::initializer_list<int>
decltype(auto) x6d = { 1, 2 }; // error, { 1, 2 } is not an expression
auto          *x7a = &i;       // decltype(x7a) is int*
decltype(auto)*x7d = &i;       // error, declared type is not plain decltype(auto)

 — end example ]

If the init-declarator-list contains more than one init-declarator, they shall all form declarations of variables. The type of each declared variable is determined as described above, and if the type that replaces the placeholder type is not the same in each deduction, the program is ill-formed.

[ Example:

auto x = 5, *y = &x;        // OK: auto is int
auto a = 5, b = { 1, 2 };   // error: different types for auto

 — end example ]

If a function with a declared return type that contains a placeholder type has multiple return statements, the return type is deduced for each return statement. If the type deduced is not the same in each deduction, the program is ill-formed.

If a function with a declared return type that uses a placeholder type has no return statements, the return type is deduced as though from a return statement with no operand at the closing brace of the function body. [ Example:

auto  f() { } // OK, return type is void
auto* g() { } // error, cannot deduce auto* from void()

 — end example ]

If the type of an entity with an undeduced placeholder type is needed to determine the type of an expression, the program is ill-formed. Once a return statement has been seen in a function, however, the return type deduced from that statement can be used in the rest of the function, including in other return statements. [ Example:

auto n = n;            // error, n's type is unknown
auto f();
void g() { &f; }       // error, f's return type is unknown
auto sum(int i) {
  if (i == 1)
    return i;          // sum's return type is int
  else
    return sum(i-1)+i; // OK, sum's return type has been deduced
}

 — end example ]

Return type deduction for a function template with a placeholder in its declared type occurs when the definition is instantiated even if the function body contains a return statement with a non-type-dependent operand. [ Note: Therefore, any use of a specialization of the function template will cause an implicit instantiation. Any errors that arise from this instantiation are not in the immediate context of the function type and can result in the program being ill-formed.  — end note ] [ Example:

template <class T> auto f(T t) { return t; }  // return type deduced at instantiation time
typedef decltype(f(1)) fint_t;                // instantiates f<int> to deduce return type
template<class T> auto f(T* t) { return *t; }
void g() { int (*p)(int*) = &f; }             // instantiates both fs to determine return types,
                                              // chooses second

 — end example ]

Redeclarations or specializations of a function or function template with a declared return type that uses a placeholder type shall also use that placeholder, not a deduced type. [ Example:

auto f();
auto f() { return 42; } // return type is int
auto f();               // OK
int f();                // error, cannot be overloaded with auto f()
decltype(auto) f();     // error, auto and decltype(auto) don't match

template <typename T> auto g(T t) { return t; } // #1
template auto g(int);                           // OK, return type is int
template char g(char);                          // error, no matching template
template<> auto g(double);                      // OK, forward declaration with unknown return type

template <class T> T g(T t) { return t; } // OK, not functionally equivalent to #1
template char g(char);                    // OK, now there is a matching template
template auto g(float);                   // still matches #1

void h() { return g(42); } // error, ambiguous

template <typename T> struct A {
  friend T frf(T);
};
auto frf(int i) { return i; } // not a friend of A<int>

 — end example ]

A function declared with a return type that uses a placeholder type shall not be virtual ([class.virtual]).

An explicit instantiation declaration ([temp.explicit]) does not cause the instantiation of an entity declared using a placeholder type, but it also does not prevent that entity from being instantiated as needed to determine its type. [ Example:

template <typename T> auto f(T t) { return t; }
extern template auto f(int); // does not instantiate f<int>
int (*p)(int) = f;           // instantiates f<int> to determine its return type, but an explicit
                             // instantiation definition is still required somewhere in the program

 — end example ]