constexpr not working if the function is declared inside class scope

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Yes, it is ill-formed. Here’s why:

A constexpr function needs to be defined (not just declared) before being used in a constant expression.

So for example:

constexpr int f(); // declare f
constexpr int x = f(); // use f - ILLEGAL, f not defined
constexpr int f() { return 5; } // define f, too late

function definitions inside a class specifier (as well as initializers and default parameters) are essentially parsed in an order like they were defined outside the class.

So this:

struct X {
  constexpr static int size() { return 5; }
  static const int array[size()]; 
};

Is parsed in this order:

struct X {
   constexpr inline static int size(); // function body defered
   static const int array[size()];  // <--- POINT A
};

constexpr inline int X::size() { return 5; }

That is, parsing of function bodies are defered until after the class specifier.

The purpose of this deferral of function body parsing is so that function bodies can forward reference class members not yet declared at that point, and also so they can use their own class as a complete type:

struct X
{
    void f() { T t; /* OK */ }
    typedef int T;
};

Compared to at namespace scope:

void f() { T t; /* error, T not declared */ }
typedef int T;

At POINT A, the compiler doesn’t have the definition of size() yet, so it can’t call it. For compile-time performance constexpr functions need to be defined ahead of their use in the translation unit before being called during compile, otherwise the compiler would have to make a multiple passes just to “link” constant expressions for evaluation.

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