9 Declarations [dcl.dcl]

9.2 Specifiers [dcl.spec]

9.2.9 Type specifiers [dcl.type] Placeholder type specifiers [dcl.spec.auto] General [dcl.spec.auto.general]

A placeholder-type-specifier designates a placeholder type that will be replaced later by deduction from an initializer.
A placeholder-type-specifier of the form type-constraint auto can be used as a decl-specifier of the decl-specifier-seq of a parameter-declaration of a function declaration or lambda-expression and, if it is not the auto type-specifier introducing a trailing-return-type (see below), is a generic parameter type placeholder of the function declaration or lambda-expression.
[Note 1: 
Having a generic parameter type placeholder signifies that the function is an abbreviated function template ([dcl.fct]) or the lambda is a generic lambda ([expr.prim.lambda]).
— end note]
A 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 trailing-return-type specifies the declared return type of the function.
Otherwise, the function declarator shall declare a function.
If the declared return type of the function contains a placeholder type, the return type of the function is deduced from non-discarded return statements, if any, in the body of the function ([stmt.if]).
The type of a variable declared using a placeholder type is deduced from its initializer.
This use is allowed in an initializing declaration ([dcl.init]) of a variable.
The placeholder type 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 declarators, each of which shall be followed by a non-empty initializer.
[Example 1: 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]
The auto type-specifier can also be used to introduce a structured binding declaration ([dcl.struct.bind]).
The auto type-specifier can also be used as the simple-type-specifier in an explicit type conversion (functional notation) ([expr.type.conv]).
A program that uses a placeholder type in a context not explicitly allowed in [dcl.spec.auto] is ill-formed.
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 by placeholder type deduction, and if the type that replaces the placeholder type is not the same in each deduction, the program is ill-formed.
[Example 2: 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 non-discarded return statements, the return type is deduced for each such 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 non-discarded 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 3: auto f() { } // OK, return type is void auto* g() { } // error: cannot deduce auto* from void() — end example]
An exported function with a declared return type that uses a placeholder type shall be defined in the translation unit containing its exported declaration, outside the private-module-fragment (if any).
[Note 2: 
The deduced return type cannot have a name with internal linkage ([basic.link]).
— end note]
If a variable or function with an undeduced placeholder type is named by an expression ([basic.def.odr]), the program is ill-formed.
Once a non-discarded 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 4: auto n = n; // error: n's initializer refers to n 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 templated function 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 3: 
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 ([temp.deduct]).
— end note]
[Example 5: 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]
If a function or function template F has a declared return type that uses a placeholder type, redeclarations or specializations of F shall use that placeholder type, not a deduced type; otherwise, they shall not use a placeholder type.
[Example 6: auto f(); auto f() { return 42; } // return type is int auto f(); // OK int f(); // error: auto and int don't match 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> extern int v; auto v = 17; // OK, redeclares v struct S { static int i; }; auto S::i = 23; // OK — end example]
A function declared with a return type that uses a placeholder type shall not be virtual ([class.virtual]).
A function declared with a return type that uses a placeholder type shall not be a coroutine ([dcl.fct.def.coroutine]).
An explicit instantiation declaration 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 7: 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]