-3
template<typename T>
void foo(T&& a)
{
cout << is_rvalue_reference<T>::value << endl;
}
struct O
{
};
O o;
foo(o); //T is deduced to o&,a is O&
foo(std::move(o)); //T is deduced to O,a is O&&
大家好。 有沒有辦法讓foo輸出1(T推導爲O & &)?關於完美轉發時的類型扣除