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placeholder type specifiers (since C++11)

From cppreference.com
< cpp‎ | language

For variables, specifies that the type of the variable that is being declared will be automatically deduced from its initializer.

For functions, specifies that the return type will be deduced from its return statements.

(since C++14)

For non-type template parameters, specifies that the type will be deduced from the argument.

(since C++17)

Contents

[edit] Syntax

auto (1) (since C++11)
decltype(auto) (2) (since C++14)
type-constraint auto (3) (since C++20)
type-constraint decltype(auto) (4) (since C++20)
type-constraint - a concept name, optionally qualified, optionally followed by a template argument list enclosed in <>
1,3) type is deduced using the rules for template argument deduction.
2,4) type is decltype(e), where e is the initializer.

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)

[edit] Explanation

A placeholder type specifier may appear in the following contexts:

  • in the type specifier of a variable: auto x = expr;. The type is deduced from the initializer.
    If the placeholder type specifier is auto or type-constraint auto (since C++20), the variable type is deduced from the initializer using the rules for template argument deduction from a function call (see template argument deduction#Other contexts for details).
    For example, given const auto& i = expr;, the type of i is exactly the type of the argument u in an imaginary template template<class U> void f(const U& u) if the function call f(expr) was compiled. Therefore, auto&& may be deduced either as an lvalue reference or rvalue reference according to the initializer, which is used in range-based for loop.

    If the placeholder type specifier is decltype(auto) or type-constraint decltype(auto) (since C++20), the deduced type is decltype(e), where e is the initializer.

    (since C++14)

    If the placeholder type specifier is used to declare multiple variables, the deduced types must match. For example, the declaration auto i = 0, d = 0.0; is ill-formed, while the declaration auto i = 0, *p = &i; is well-formed and the auto is deduced as int.

  • in the type-id of a new expression. The type is deduced from the initializer. For new T init (where T contains a placeholder type, init is either a parenthesized initializer or a brace-enclosed initializer list), the type of T is deduced as if for variable x in the invented declaration T x init;.
  • (since C++14) in the return type of a function or lambda expression: auto& f();. The return type is deduced from the operand of its non-discarded (since C++17) return statement.
    See function#Return_type_deduction.
  • (since C++17) in the parameter declaration of a non-type template parameter: template<auto I> struct A;. Its type is deduced from the corresponding argument.

Furthermore, auto and type-constraint auto (since C++20) can appear in:

(since C++14)

If type-constraint is present, let T be the type deduced for the placeholder, T must satisfy the immediately-declared constraint of type-constraint. That is,

  • If type-constraint is Concept<A1, ..., An>, then the constraint expression Concept<T, A1, ..., An> must be valid and return true;
  • otherwise (type-constraint is Concept without an argument list), the constraint expression Concept<T> must be valid and return true.
(since C++20)

[edit] Notes

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

Mixing auto variables and functions in one declaration, as in auto f() -> int, i = 0; is not allowed.

The auto specifier may also be used with a function declarator that is followed by a trailing return type, in which case the declared return type is that trailing return type (which may again be a placeholder type).

auto (*p)() -> int; // declares p as pointer to function returning int
auto (*q)() -> auto = p; // declares q as pointer to function returning T
                         // where T is deduced from the type of p

The auto specifier may also be used in a structured binding declaration.

(since C++17)

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)

[edit] 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 << "a, before modification through c2 = " << a << '\n';
    ++c2;
    std::cout << "a, after modification through c2 = " << 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 C++17, std::initializer_list<int> before
    auto m{5};       // OK: type of m is int as of C++17, 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++
}

Possible output:

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