Cannot construct constexpr array from braced-init-list

Question

I've implemented a constexpr array like this:

template <typename T>
class const_array {
  const T* p;
  unsigned n;
public:
  template <unsigned N>
  constexpr const_array(const T(&a)[N]): p(a), n(N) { }

  constexpr unsigned size() const { return n; }
};

int main(int argc, char* argv[]) {
  // works
  static_assert(const_array<double>{{1.,2.,3.}}.size() == 3);

  // doesn't compile
  constexpr const_array<double> a{{1.,2.,3.}};
  static_assert(a.size() == 3);
}

Why is it that the first static_assert compiles, but initializing a fails?I'm using gcc 6.2.0. I'm getting

: In function 'int main(int, char**)':
: error: 'const_array<double>{((const double*)(&<anonymous>)), 3u}' is not a constant expression
   constexpr const_array<double> a{{1.,2.,3.}};
                                        ^
test/const_array.cc:17:3: error: non-constant condition for static assertion
   static_assert(a.size() == 3);
   ^~~~~~~~~~~~~

Answer 1

The compiler is complaining that the initializer of a.p is not a constant expression. It's failing §5.20/5.2:

if the value is of pointer type, it contains the address of an object with static storage duration, the address past the end of such an object (5.7), the address of a function, or a null pointer value

In other words, only pointer values known to the linker are valid constants. (Also, in your example the pointer is dangling.)

The first static_assert doesn't trip this because p is discarded and the value of n is a constant expression. Constant expressions may have non-constant subexpressions.

This works:

static constexpr double arr[] = { 1.,2.,3. };
constexpr const_array<double> a{ arr };
static_assert( a.size() == 3 );

Credit to @Jarod42 for pointing out the issue in the comments.