int i = 42; decltype((i)) d; //error: d is int& and must be initialized decltype(i) e; //ok: e is an (uninitialized) int

下面是引自 《C++ primer 5th edition》的一段话：在2.5 Dealing with types的“decltype and reference” 这一小节里面

If we wrap the variable’s name in one or more sets of parentheses, the compiler will evaluate the operand as an expression. A variable is an expression that can be the left-hand side of an assignment. As a result, decltype on such an expression yields a reference

不能理解为什么 left-hand side of an assignment 会使得decltype最后yield一个引用。

求解释，万分感谢！