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

First erase the outermost decltype():

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

In other words, the first decltype corresponds to (i), and the second one handles i.

In C++ expressions, the first one is "lvalue expression"; the second one is "a variable".
Because it is lvalue, decltype will be understood as a reference.

Answer 2

There is this sentence earlier in the book:

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 that yield objects that can stand on the left- hand side of the assignment

So decltype when applied to an lvalue will return a reference type

Answer 3

Because the type of (i) is a reference.
decltype(var) is the type of this variable; decltype(expr) is the type of this expression.