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# auto specifier (since C++11)

From cppreference.com
< cpp‎ | language

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 is a trailing return type or will be deduced from its return statements (since C++14).

### [edit] Syntax

auto variable initializer | (1) | (since C++11)
auto function -> return type | (2) | (since C++11)
auto function | (3) | (since C++14)
decltype(auto) variable initializer | (4) | (since C++14)
decltype(auto) function | (5) | (since C++14)

### [edit] Explanation

1) When declaring variables in block scope, in namespace scope, in init statements of for loops, etc, the keyword auto may be used as the type specifier.
Once the type of the initializer has been determined, the compiler determines the type that will replace the keyword auto using the rules for template argument deduction from a function call (see template argument deduction#Other contexts for details). The keyword auto may be accompanied by modifiers, such as const or &, which will participate in the type deduction. 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.
2) In a function declaration that uses the trailing return type syntax, the keyword auto does not perform automatic type detection. It only serves as a part of the syntax.
3) In a function declaration which does not use the trailing return type syntax, the keyword auto indicates that the return type will be deduced from the operand of its return statement using the rules for template argument deduction.
4) If the declared type of the variable is decltype(auto), the keyword auto is replaced with its initializer, and the actual type is deduced using the rules for decltype.
5) If the return type of the function is declared decltype(auto), the keyword auto is replaced with the operand of its return statement, and the actual return type is deduced using the rules for decltype.

### [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.

### [edit] Example

the example showing output using one of the implementations where typeinfo::name prints full type names; filter through c++filt -t if using gcc or similar

Run this code

    #include <iostream>
    #include <cmath>
    #include <typeinfo>

    template<class T, class U>
    auto add(T t, U u) -> decltype(t + u) // the return type is the type of operator+(T,U)
    {
        return t + u;
    }

    auto get_fun(int arg)->double(*)(double) // same as double (*get_fun(int))(double)
    {
        switch (arg) {
            case 1: return std::fabs;
            case 2: return std::sin;
            default: return std::cos;
        }
    }

    int main()
    {
        auto a = 1 + 2;
        std::cout << "type of a: " << typeid(a).name() << '\n';
        auto b = add(1, 1.2);
        std::cout << "type of b: " << typeid(b).name() << '\n';
        //auto int c; //compile-time error
        auto d = {1, 2};
        std::cout << "type of d: " << typeid(d).name() << '\n';

        auto my_lambda = [](int x) { return x + 3; };
        std::cout << "my_lambda: " << my_lambda(5) << '\n';

        auto my_fun = get_fun(2);
        std::cout << "type of my_fun: " << typeid(my_fun).name() << '\n';
        std::cout << "my_fun: " << my_fun(3) << '\n';
    }

Possible output:

    type of a: int
    type of b: double
    type of d: std::initializer_list<int>
    my_lambda: 8
    type of my_fun: double (*)(double)
    my_fun: 0.14112

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