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# std::declval

From cppreference.com
< cpp‎ | utility

Defined in header `<utility>` |  |
template< class T > typename std::add_rvalue_reference<T>::type declval(); |  | (since C++11)

Converts any type `T` to a reference type, making it possible to use member functions in decltype expressions without specifying constructors.

`declval` is commonly used in templates where acceptable template parameters may have no constructor in common, but have the same member function whose return type is needed. Note that because no definition exists for `declval`, it can only be used in unevaluated contexts; it is an error to evaluate an expression that contains this function.

### [edit] Parameters

(none)

### [edit] Return value

Cannot be called and thus never returns a value. The return type is `T&&` unless `T` is an lvalue reference type, in which case the return type is `T&`.

### [edit] Exceptions

`noexcept` specification:
noexcept

### [edit] Example

Run this code

    #include <utility>
    #include <iostream>

    struct Default {
        int foo() const {return 1;}
    };

    struct NonDefault {
        NonDefault(const NonDefault&) {}
        int foo() const {return 1;}
    };

    int main()
    {
        decltype(Default().foo()) n1 = 1; // int n1
    //  decltype(NonDefault().foo()) n2 = n1; // will not compile
        decltype(std::declval<NonDefault>().foo()) n2 = n1; // int n2
        std::cout << "n2 = " << n2 << '\n';
    }

Output:

    n2 = 1

### [edit] See also

decltype specifier | defines a type equivalent to the type of an expression (C++11) [edit]
result_of (C++11) | deduces the return type of a function call expression (class template) [edit]

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