decltype and parentheses
C++C++11Type InferenceDecltypeC++ Problem Overview
I don't understand the last line of the example on page 148 of the FCD (§7.6.1.2/4):
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&
Why do the parentheses make a difference here? Shouldn't it simply be double
like in the line above?
C++ Solutions
Solution 1 - C++
Just above that example, it says
> - if e is an unparenthesized id-expression or a class member access (5.2.5), decltype(e) is the type of the entity named by e. > - if e is an lvalue, decltype(e) is T&, where T is the type of e;
I think decltype(a->x)
is an example of the "class member access" and decltype((a->x))
is an example of lvalue.
Solution 2 - C++
decltype(a->x)
This gives you the type of the member variable A::x
, which is double
.
decltype((a->x))
This gives you the type of the expression (a->x)
, which is an lvalue expression (hence why it is a const reference--a
is a const A*
).
Solution 3 - C++
The added parens are turning it into a lvalue.
MSDN says
The inner parentheses cause the statement to be evaluated as an expression instead of a member access. And because a is declared as a const pointer, the type is a reference to const double.