C ++ rvalue引用了对左值引用的赋值

Question

让我们假设：

template<typename T>
T&& f(T&& t) {
  return std::forward<T>(t);
}

Presenter p;

auto x = f(std::move(p));   // x is new Presenter object - move ctor invoked
auto x = f(p);                // x is new Presenter object - copy ctor invoked
auto& x = f(p);             // x is lvalueRef, no copy / move ctor invoked
auto&& x = f(p);       // x is lvalueRef, no copy /move ctor invoked
auto&& x = f(std::move(p));   // x is rvalueRef, no copy / move ctor invoked

这很好，对于那种情况而言：

auto& x = f(std::move(p));  // x is lvalueRef, no copy / move ctor invoked

为什么我们能够协助lvalueRef rvalueRef？

例如我们不能这样做：

auto& c = 1;

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