Placeholder type specifiers (since C++11)
A placeholder type specifier designates a placeholder type that will be replaced later, typically by deduction from an initializer.
Syntax
type-constraint (optional) auto
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(1) | ||||||||
type-constraint (optional) decltype(auto)
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(2) | (since C++14) | |||||||
type-constraint | - | (since C++20) a concept name, optionally qualified, optionally followed by a template argument list enclosed in <>
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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
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 declarationsIn the following parameter declarations, the type of the parameter declared can be of syntax (1):
|
(since C++14) |
|
(since C++17) |
|
(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 int auto g() { return 0.0; } // OK since C++14: g returns double auto 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 specifiers 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 f() -> int; auto (*fp)() -> auto = f; // OK: the “auto” in the trailing return type // can be deduced from f
Structured binding declarationsThe 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 castThe 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
Example
#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 reference template<class F, class... Args> decltype(auto) PerfectForward(F fun, Args&&... args) { return fun(std::forward<Args>(args)...); } template<auto n> // C++17 auto parameter declaration auto 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 int auto 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 a std::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> before auto 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 expressions auto 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]
- 10.1.7.4 The
- C++14 standard (ISO/IEC 14882:2014):
- 7.1.6.4
auto
specifier [dcl.spec.auto]
- 7.1.6.4
- C++11 standard (ISO/IEC 14882:2011):
- 7.1.6.4
auto
specifier [dcl.spec.auto]
- 7.1.6.4