9 Declarations [dcl.dcl]

9.2 Specifiers [dcl.spec]

9.2.8 Type specifiers [dcl.type]

9.2.8.5 Placeholder type specifiers [dcl.spec.auto]

placeholder-type-specifier:
	type-constraint auto
	type-constraint decltype ( auto )
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 in the decl-specifier-seq of a parameter-declaration of a function declaration or lambda-expression and signifies that the function is an abbreviated function template ([dcl.fct]) or the lambda is a generic lambda ([expr.prim.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 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.
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
 ]
The auto type-specifier can also be used to introduce a structured binding declaration ([dcl.struct.bind]).
A placeholder type can also be used in the type-specifier-seq in the new-type-id or type-id of a new-expression and as a decl-specifier of the parameter-declaration's decl-specifier-seq in a template-parameter.
A program that uses a placeholder type in a context not explicitly allowed in this subclause 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
:
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
:
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
:
There is no restriction on the linkage of the deduced return type.
— end note
 ]
If the name of an entity with an undeduced placeholder type appears in an expression, 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
:
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 entity that is a function or 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 ([temp.deduct]).
— 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.
Similarly, redeclarations or specializations of a function or function template with a declared return type that does not use a placeholder type shall not use a placeholder.
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>
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.
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
:
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
 ]

9.2.8.5.1 Placeholder type deduction [dcl.type.auto.deduct]

Placeholder type deduction is the process by which a type containing a placeholder type is replaced by a deduced type.
A type T containing a placeholder type, and a corresponding initializer e, are determined as follows:
  • for a non-discarded return statement that occurs in a function declared with a return type that contains a placeholder type, T is the declared return type and e is the operand of the return statement. If the return statement has no operand, then e is void();
  • for a variable declared with a type that contains a placeholder type, T is the declared type of the variable and e is the initializer. If the initialization is direct-list-initialization, the initializer shall be a braced-init-list containing only a single assignment-expression and e is the assignment-expression;
  • for a non-type template parameter declared with a type that contains a placeholder type, T is the declared type of the non-type template parameter and e is the corresponding template argument.
In the case of a return statement with no operand or with an operand of type void, T shall be either type-constraint decltype(auto) or cv type-constraint auto.
If the deduction is for a return statement and e is a braced-init-list ([dcl.init.list]), the program is ill-formed.
If the placeholder-type-specifier is of the form type-constraint auto, the deduced type replacing T is determined using the rules for template argument deduction.
Obtain P from T by replacing the occurrences of type-constraint auto either with a new invented type template parameter U or, if the initialization is copy-list-initialization, with std​::​initializer_­list<U>.
Deduce a value for U using the rules of template argument deduction from a function call, where P is a function template parameter type and the corresponding argument is e.
If the deduction fails, the declaration is ill-formed.
Otherwise, 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
auto x3{ 1, 2 };                // error: not a single element
auto x4 = { 3 };                // decltype(x4) is std​::​initializer_­list<int>
auto x5{ 3 };                   // decltype(x5) is int
— 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-type-specifier is of the form type-constraint decltype(auto), T shall be the placeholder alone.
The type deduced for T is determined as described in [dcl.type.simple], as though e had been the operand of the decltype.
Example
:
int i;
int&& f();
auto           x2a(i);          // decltype(x2a) is int
decltype(auto) x2d(i);          // decltype(x2d) is int
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
 ]
For a placeholder-type-specifier with a type-constraint, the immediately-declared constraint ([temp.param]) of the type-constraint for the type deduced for the placeholder shall be satisfied.