Function initialization with int& argument by int&&

Question

I have following code:

#include <type_traits>

struct TType
{
    int a = 0;

    bool operator() (int&)
    {
        return true;
    }
};

int main()
{
    static_assert(std::is_same<decltype(std::declval<TType>()(std::declval<int>())), bool>::value, "wtf?");
    return 0;
}

If I try to compile it with g++-4.8.2 then I receive an error:

main.cpp:321:82: error: no match for call to ‘(JetPlane) (int)’
static_assert(std::is_same<decltype(std::declval<JetPlane>()(std::declval<int>())), bool>::value, "wtf?");
                                                                                ^
main.cpp:265:8: note: candidate is:
struct JetPlane
       ^
main.cpp:279:7: note: bool JetPlane::operator()(int&)
bool operator() (int&)
     ^
main.cpp:279:7: note:   no known conversion for argument 1 from ‘int’ to ‘int&’

I don't understand note: no known conversion for argument 1 from ‘int’ to ‘int&’ line. So the question is: why g++ interprets return type of std::declval<int>() like int and not line int&& though std::declval declaration looks like:

template< class T >
typename std::add_rvalue_reference<T>::type declval();

I understand that it's prohibited to bind int&& to int. But why then compiler doesn't print: note: no known conversion for argument 1 from ‘int’ to ‘int&’ line. May be I don't understand something and compiler changes somehow return type of std::declval<int>() from int&& on int in std::declval<TType>()(std::declval<int>())?

Thank you for help!

Answer 1

The problem is, that you cannot bind an xvalue to a non-const lvalue reference.

Let's look at the expression

std::declval<int>()

The return type of std::declval<int> is indeed int&&. Hence the above expression is an xvalue expression of type int. Note that in C++ an expression never has a reference type.

But your operator

bool operator() (int&)

takes its argument by non-const lvalue referenece. If you change the perameter type to const int&, i.e.

bool operator() (const int&)

everything should work fine.