Constexpr and std::pair

Clang Frontend
#1

Consider the following code:

  constexpr int x = 0;
  constexpr int y = 1;
  constexpr std::pair<const int &, const int &> p {x,y};

Shouldn’t that compile? (with -std=c++1y)
In C++14, we have:
         constexpr pair(const T1& x, const T2& y);
(and yes, libc++ implements it that way).

But clang rejects it, with:
junk2.cpp:18:46: error: constexpr variable 'p' must be initialized by a constant
      expression
        constexpr std::pair<const int&, const int&> p { x, y };
                                                    ^~~~~~~~
junk2.cpp:18:46: note: reference to 'x' is not a constant expression
junk2.cpp:15:16: note: declared here
        constexpr int x = 0;

with the caret pointing at the ‘p’ in "> p { x, y };”

Followup question. Assuming that this should compile, then what about this?

  typedef std::pair<const int &, const int &> p_t;
  constexpr int x = 0;
  constexpr int y = 1;
  constexpr p_t foo () { return p_t{x,y}; }

  constexpr p_t bar = foo();

pair’s copy and copy constructors are defined as “= default”.

But, 20.3.2/2 says:

The defaulted move and copy constructor, respectively, of pair shall be a constexpr function if and only if all required element-wise initializations for copy and move, respectively, would satisfy the requirements for a constexpr function.

— Marshall

#2

Consider the following code:

        constexpr int x = 0;
        constexpr int y = 1;
        constexpr std::pair<const int &, const int &> p {x,y};

Shouldn't that compile? (with -std=c++1y)

Depends. If this is namespace-scope, then yes. If it's block-scope, then no
(it should compile only if the addresses of x and y are constant).

In C++14, we have:
         constexpr pair(const T1& x, const T2& y);
(and yes, libc++ implements it that way).

But clang rejects it, with:
junk2.cpp:18:46: error: constexpr variable 'p' must be initialized by a
constant
      expression
        constexpr std::pair<const int&, const int&> p { x, y };
                                                    ^~~~~~~~
junk2.cpp:18:46: note: reference to 'x' is not a constant expression
junk2.cpp:15:16: note: declared here
        constexpr int x = 0;

with the caret pointing at the 'p' in "> p { x, y };"

Followup question. Assuming that this should compile, then what about this?

        typedef std::pair<const int &, const int &> p_t;
        constexpr int x = 0;
        constexpr int y = 1;
        constexpr p_t foo () { return p_t{x,y}; }

        constexpr p_t bar = foo();

pair's copy and copy constructors are defined as "= default".

But, 20.3.2/2 says:

The defaulted move and copy constructor, respectively, of pair shall be a
constexpr function if and only if all required element-wise initializations
for copy and move, respectively, would satisfy the requirements for a
constexpr function.

I think this should work (and indeed it seems to with libc++, but not with
libstdc++ where the pair move constructor is not constexpr).