What expressions yield a reference type when decltype is applied to them?

Question

I was reading C++ Primer and couldn\'t quite understand when an expression yields an object type, and when it yields a reference type to the object.

I quote from the

Answer 1

For expressions, as in your examples decltype will provide a reference type if the argument is lvalue.

7.1.6.2p4:

The type denoted by decltype(e) is defined as follows:
  — if e is an unparenthesized id-expression or an unparenthesized class member access (5.2.5), decltype(e)     is the type of the entity named by e. If there is no such entity, or if e names a set of overloaded functions,     the program is ill-formed;
  — otherwise, if e is an xvalue, decltype(e) is T&&, where T is the type of e;
  — otherwise, if e is an lvalue, decltype(e) is T&, where T is the type of e;
  — otherwise, decltype(e) is the type of e.
The operand of the decltype specifier is an unevaluated operand (Clause 5).
[ Example:
const int&& foo();
int i;
struct A { double x; };
const A* a = new A();
decltype(foo()) x1 = i; // type is const int&&
decltype(i) x2; // type is int
decltype(a->x) x3; // type is double
decltype((a->x)) x4 = x3; // type is const double&
—end example ]