C++ decltype((i))为什么是个引用？

Question

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！

Answer 1

先把最外面的 decltype() 給抹掉：

• decltype((i)) -> (i)
• decltype(i) -> i

也就是說，第一個decltype是對應(i)，第二個處理i。

在C++表達式中，第一個是"lvalue表達式"；第二個是"一個變數"。
因為是lvalue，因此decltype會理解為引用。

Answer 2

書中稍早一點有這樣一句話：

As we'll see in § 4.1.1 (p. 135), some expressions will cause decltype to yield a reference type. Generally speaking, decltype returns a reference type for expressions type. Generally speaking, decltype returns a reference type for expressions that yield objects that the stects that yield that that yield-that that yield-that that yield that that yield-that that yield that that a hand side of the assignment

所以decltype作用在左值上會回傳引用型別

Answer 3

因為(i)的型別就是個引用啊。
decltype(var)就是這個變數的型別；decltype(expr)就是這個表達式的型別。