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]

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.