Is there a reason declval returns add_rvalue_reference instead of add_lvalue_reference

Question

changing a type into a reference to a type, allows one to access the members of the type without creating an instance of the type. This seems to be true for bot

Answer 1

Yes, the use of add_rvalue_reference gives the client the choice of specifying whether he wants an lvalue or rvalue object of the given type:

#include 
#include 
#include 
#ifndef _MSC_VER
#   include 
#endif
#include 
#include 
#include 

template 
std::string
type_name()
{
    typedef typename std::remove_reference::type TR;
    std::unique_ptr own
           (
#ifndef _MSC_VER
                abi::__cxa_demangle(typeid(TR).name(), nullptr,
                                           nullptr, nullptr),
#else
                nullptr,
#endif
                std::free
           );
    std::string r = own != nullptr ? own.get() : typeid(TR).name();
    if (std::is_const::value)
        r += " const";
    if (std::is_volatile::value)
        r += " volatile";
    if (std::is_lvalue_reference::value)
        r += "&";
    else if (std::is_rvalue_reference::value)
        r += "&&";
    return r;
}

int
main()
{
    std::cout << type_name())>() << '\n';
    std::cout << type_name())>() << '\n';
}

Which for me outputs:

int&&
int&