-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。 有没有办法让foo输出1(T推导为O & &)?关于完美转发时的类型扣除