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
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 ]