Initialization of constexpr member variable using constexpr member function

Question

I wanted to initialize a constexpr member variable using a constexpr member function but it didn't compile. It was OK when I moved the function out of the class. Why does it happen? Is there any way to use class member constexpr functions to initialize member constexpr variables?

I'm using Apple LLVM version 8.0.0 (clang-800.0.38).

Thanks for any help.

constexpr static int Add_Ext(int a, int b) { return a + b; }


class Foo
{
public:
    constexpr static int Add_InClass(int a, int b) { return a + b; }

    // This is OK.
    constexpr static int kConstantX = Add_Ext(1, 2);

    // This results in a compile error.
    constexpr static int kConstantY = Add_InClass(1, 2);

};

clang error message:

Constexpr variable 'kConstantY' must be initialized by a constant expression

Answer 1

From CWG-1255 and CWG-1626

The Standard should make clear that a constexpr member function cannot be used in a constant expression until its class is complete.

It seems like your code is intentionally unacceptable. However, the standard should make it more clear in that matter.

Is there any way to use class member constexpr functions to initialize member constexpr variables?

w.r.t. those DRs, No. You can move it to outside or to another class instead.

Answer 2

Thanks to Danh. I looked into the WG thread you told, and after all, I found I can use member constexpr functions if I make it a class template.

// This works (it doesn't if you make it as a non-template class).
template <typename T>
class TemplateFoo
{
public:
    constexpr static T Add_InClass(T a, T b) { return a + b; }
    constexpr static T kConstantY = Add_InClass(1, 2);

};

// Even such a meaningless template works too. 
template <typename T>
class TemplateBar
{
public:
    constexpr static int Add_InClass(int a, int b) { return a + b; }
    constexpr static int kConstantY = Add_InClass(1, 2);

};

// So, this would be a (dirty but) useful workaround 
// when you don't want to use the class as a template.
// Any type could be used as the template argument, which has no meaning.
using Bar = TemplateBar<char>;