Search code examples
c++c++11type-inferencedecltype

What does decltype((...)) with double parentheses mean?

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?

Solution

  • 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.