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# decltype specifier

From cppreference.com
< cpp‎ | language

Inspects the declared type of an entity or queries the return type of an expression.

### [edit] Syntax

`decltype (` entity `)` | (1) | (since C++11)
`decltype (` expression `)` | (2) | (since C++11)

### [edit] Explanation

1) If the argument is either the unparenthesised name of an object/function, or is a member access expression (`object.member` or `pointer->member`), then the decltype specifies the declared type of the entity specified by this expression.
2) If the argument is any other expression of type `T`, then
a) if the value category of expression is xvalue, then the decltype specifies `T&&`
b) if the value category of expression is lvalue, then the decltype specifies `T&`
c) otherwise, decltype specifies `T`

If expression is a function call which returns a prvalue of class type or is a comma expression whose right operand is such a function call, a temporary object is not introduced for that prvalue. The class type need not be complete or have an available destructor. This rule doesn't apply to sub-expressions:in decltype(f(g())), g() must have a complete type, but f() need not.

Note that if the name of an object is parenthesised, it becomes an lvalue expression, thus decltype(arg) and decltype((arg)) are often different types.

`decltype` is useful when declaring types that are difficult or impossible to declare using standard notation, like lambda-related types or types that depend on template parameters.

### [edit] Keywords

`decltype`

### [edit] Example

Run this code

    #include <iostream>

    struct A {
       double x;
    };
    const A* a = new A{0};

    decltype( a->x ) x3;       // type of x3 is double (declared type)
    decltype((a->x)) x4 = x3;  // type of x4 is const double& (lvalue expression)

    template <class T, class U>
    auto add(T t, U u) -> decltype(t + u); // return type depends on template parameters

    int main()
    {
        int i = 33;
        decltype(i) j = i*2;

        std::cout << "i = " << i << ", "
                  << "j = " << j << '\n';

        auto f = [](int a, int b) -> int {
           return a*b;
        };

        decltype(f) f2 = f; // the type of a lambda function is unique and unnamed
        i = f(2, 2);
        j = f2(3, 3);

        std::cout << "i = " << i << ", "
                  << "j = " << j << '\n';
    }

Output:

    i = 33, j = 66
    i = 4, j = 9

### [edit] See also

declval (C++11) | obtains the type of expression in unevaluated context (function template) [edit]

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