Placeholder type specifiers (since C++11)

 

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A placeholder type specifier designates a placeholder type that will be replaced later, typically by deduction from an initializer.

Syntax

type-constraint (optional)auto	(1)
type-constraint (optional)decltype(auto)	(2)	(since C++14)

type-constraint

-

(since C++20) a concept name, optionally qualified, optionally followed by a template argument list enclosed in <>

1) Type is deduced using the rules for template argument deduction.

2) Type is decltype(expr), where expr is the initializer or the operands used in return statements.

The placeholder auto may be accompanied by modifiers, such as const or &, which will participate in the type deduction. The placeholder decltype(auto) must be the sole constituent of the declared type.(since C++14)

If type-constraint is present, let T be the type deduced for the placeholder, the type-constraint introduces a constraint expression as follows: If type-constraint is Concept<A1, ..., An>, then the constraint expression is Concept<T, A1, ..., An>; otherwise (type-constraint is Concept without an argument list), the constraint expression is Concept<T>. Deduction fails if the constraint expression is invalid or returns false.

(since C++20)

Explanation

A placeholder type specifier may appear in the following contexts:

Parameter declarations In the following parameter declarations, the type of the parameter declared can be of syntax (1): If a parameter of a lambda expression has a placeholder type, the lambda expression is a generic lambda.	(since C++14)
If a constant template parameter has a placeholder type, its type is deduced from the corresponding template argument.	(since C++17)
If a parameter of a function declaration has a placeholder type, an abbreviated function template is declared.	(since C++20)

Function declarations

A placeholder type can appear in the declaration specifiers for a function declarator that includes a trailing return type.

A placeholder type can appear in the declaration specifiers or type specifiers in the declared return type of a function declarator. Return type deduction will be applied in this case.

(since C++14)

auto f()->int;// OK: f returns intauto g(){return0.0;}// OK since C++14: g returns doubleauto h();// OK since C++14: h’s return type will be deduced when it is defined

Variable declarations

The type of a variable declared using a placeholder type is deduced from its initializer. This use is allowed in an initializing declaration of a variable.

The placeholder type can only appear as one of the declaration specifiers in the declaration specifier sequence or as one of the type specifiers in a trailing return type that specifies the type that replaces such a declaration specifier. In this case, the declaration must declare at least one variable, and each variable must have a non-empty initializer.

// “auto”s in declaration specifiersauto x =5;// OK: x has type intconstauto*v =&x, u =6;// OK: v has type const int*, u has type const intstaticauto y =0.0;// OK: y has type double   auto f()->int;auto(*fp)()->auto= f;// OK: the “auto” in the trailing return type// can be deduced from f

Structured binding declarations The auto specifier can be used in a structured binding declaration.

(since C++17)

new expressions

A placeholder type can be used in the type specifier sequence of the type-id of a new expression. In such a type-id, the placeholder type must appear as one of the type specifiers in the type specifier sequence or a trailing return type that specifies the type that replaces such a type specifier.

Function-style cast The auto type specifier can be used as the type specifier of a function-style cast.

(since C++23)

Notes

Until C++11, auto had the semantic of a storage duration specifier.

A program that uses a placeholder type in a context not explicitly stated above is ill-formed.

If a declaration declares multiple entities, and the declaration specifier sequence uses a placeholder type, the program is ill-formed if any of the following conditions is satisfied:

• Some of the entities declared are not variables.
• The type that replaces the placeholder type is not the same in each deduction.

auto f()->int, i =0;// Error: declares a function and a variable with “auto”auto a =5, b ={1, 2};// Error: different types for “auto”

The auto keyword may also be used in a nested name specifier. A nested name specifier of the form auto:: is a placeholder that is replaced by a class or enumeration type following the rules for constrained type placeholder deduction.

(concepts TS)

Feature-test macro	Value	Std	Feature
__cpp_decltype_auto	201304L	(C++14)	decltype(auto)

Keywords

auto, decltype

Example

Run this code

#include <iostream>#include <utility>   template<class T, class U>auto add(T t, U u){return t + u;}// the return type is the type of operator+(T, U)   // perfect forwarding of a function call must use decltype(auto)// in case the function it calls returns by referencetemplate<class F, class... Args> decltype(auto) PerfectForward(F fun, Args&&... args){return fun(std::forward<Args>(args)...);}   template<auto n>// C++17 auto parameter declarationauto f()->std::pair<decltype(n), decltype(n)>// auto can't deduce from brace-init-list{return{n, n};}   int main(){auto a =1+2;// type of a is intauto b = add(1, 1.2);// type of b is double static_assert(std::is_same_v<decltype(a), int>); static_assert(std::is_same_v<decltype(b), double>);   auto c0 = a;// type of c0 is int, holding a copy of a decltype(auto) c1 = a;// type of c1 is int, holding a copy of a decltype(auto) c2 =(a);// type of c2 is int&, an alias of astd::cout<<"before modification through c2, a = "<< a <<'\n';++c2;std::cout<<" after modification through c2, a = "<< a <<'\n';   auto[v, w]= f<0>();//structured binding declaration   auto d ={1, 2};// OK: type of d is std::initializer_list<int>auto n ={5};// OK: type of n is std::initializer_list<int>// auto e{1, 2}; // Error as of DR n3922, std::initializer_list<int> beforeauto m{5};// OK: type of m is int as of DR n3922, initializer_list<int> before// decltype(auto) z = { 1, 2 } // Error: {1, 2} is not an expression   // auto is commonly used for unnamed types such as the types of lambda expressionsauto lambda =[](int x){return x +3;};   // auto int x; // valid C++98, error as of C++11// auto x; // valid C, error in C++   [](...){}(c0, c1, v, w, d, n, m, lambda);// suppresses "unused variable" warnings}

Possible output:

before modification through c2, a = 3 after modification through c2, a = 4

Defect reports

The following behavior-changing defect reports were applied retroactively to previously published C++ standards.

DR	Applied to	Behavior as published	Correct behavior
CWG 1265	C++11	the auto specifier could be used to declare a function with a trailing return type and define a variable in one declaration statement	prohibited
CWG 1346	C++11	a parenthesized expression list could not be assigned to an auto variable	allowed
CWG 1347	C++11	a declaration with the auto specifier could define two variables with types T and std::initializer_list<T> respectively	prohibited
CWG 1852	C++14	the auto specifier in decltype(auto) was also a placeholder	not a placeholder in this case
CWG 1892	C++11	the return type of a function pointer type-id could be auto	prohibited
CWG 2476	C++11	the resolution of CWG issue 1892 prohibited the deduction of the return type of function pointer variables from initializers	allowed

References

• C++23 standard (ISO/IEC 14882:2024):

• 9.2.9.6 Placeholder type specifiers [dcl.spec.auto]

• C++20 standard (ISO/IEC 14882:2020):

• 9.2.8.5 Placeholder type specifiers [dcl.spec.auto]

• C++17 standard (ISO/IEC 14882:2017):

• 10.1.7.4 The auto specifier [dcl.spec.auto]

• C++14 standard (ISO/IEC 14882:2014):

• 7.1.6.4 auto specifier [dcl.spec.auto]

• C++11 standard (ISO/IEC 14882:2011):

• 7.1.6.4 auto specifier [dcl.spec.auto]