Why is a non-const rvalue move constructor called in this case?

Question

I have seen the related questions and they mostly talk about if we should have const rvalue references as a parameter or not. But I still fail to reason why a non-const move constructor is being called in the following code:

    #include <iostream>
    using namespace std;

    class A 
    {
    public:
      A (int const &&i) { cout << "const rvalue constructor"; }
      A (int &&i) { cout << "non const rvalue constructor"; }
   };


   int const foo (void)
   {
     const int i = 3;
     return i;
   }

  int main (void)
  {
     A a(foo());
  }

Answer 1

Here is a slightly modified version of your code:

#include <iostream>

#if 0
using T = int;
#else
struct T {T(int){}};
#endif

    using namespace std;
    class A {
      public:
      A (T const &&i) { cout << "const rvalue constructor"; }
      A (T &&i) { cout << "non const rvalue constructor"; }
   };


   T const
   foo (void)
   {
     const T i = 3;
     return i;
   }

  int main()
  {
    A a(foo());
  }

When T == int, you get the non-const overload. When T is a class type, you get the const overload. This behavior falls out of section 8.2.2 [expr.type]/p2:

If a prvalue initially has the type “cv T”, where T is a cv-unqualified non-class, non-array type, the type of the expression is adjusted to T prior to any further analysis.

Translation: The language doesn't have const-qualified scalar prvalues. They simply don't exist.