C++ static constexpr field with incomplete type

Question

I'm trying to compile this code, but g++ complains about ZERO having an incomplete type. Does this mean that in C++ a struct cannot contain a static constexpr instance of itself? If so, why?

struct Cursor
{
    size_t row,column;

    static constexpr Cursor ZERO {0,0};
    //error: constexpr const Cursor Cursor::ZERO has incomplete type
};

EDIT: I understand that Cursor cannot have a complete type when I declare ZERO. What I'd like to know is: is there any way I can have ZERO belonging to Cursor and still being constexpr?

Answer 1

Unfortunately, you simply cannot do this!

Some static constexpr members may be initialised inline:

[C++11 9.4.2/3]: [..] A static data member of literal type can be declared in the class definition with the constexpr specifier; if so, its declaration shall specify a brace-or-equal-initializer in which every initializer-clause that is an assignment-expression is a constant expression. [..]

Cursor is a literal type, so this counts.

And the use of Cursor itself as a static data member within its own type is not a problem, as long as you initialise it at lexical namespace scope:

[C++11: 9.4.2/2]: The declaration of a static data member in its class definition is not a definition and may be of an incomplete type other than cv-qualified void. The definition for a static data member shall appear in a namespace scope enclosing the member’s class definition. In the definition at namespace scope, the name of the static data member shall be qualified by its class name using the :: operator. The initializer expression in the definition of a static data member is in the scope of its class (3.3.7).

But you can't do that with constexpr:

[C++11: 7.1.5/9]: A constexpr specifier used in an object declaration declares the object as const. Such an object shall have literal type and shall be initialized. [..]

I think all of this wording could be improved but, in the meantime, I think you're going to have to make ZERO a non-member in the enclosing namespace.

Answer 2

is there any way I can have ZERO belonging to Cursor and still being constexpr?

Yes, if you count nested subclasses as "belonging to" the containing class:

struct Cursor
{
    size_t row,column;

    struct Constants;
};

struct Cursor::Constants
{
    static constexpr Cursor ZERO {0,0};
};

Answer 3

This answer to a similar question revealed that this is, in fact, possible to accomplish. You just have to put the constexpr keyword with the definition rather than the declaration:

#include <iostream>
#include <array>

struct Cursor
{
    static Cursor const ZERO;
    std::size_t row, column;
};

constexpr Cursor const Cursor::ZERO{ 0, 0 };

int main(int, char**) noexcept
{
    // using the values in a template argument ensure compile-time usage
    std::array<int, Cursor::ZERO.row> row_arr{};
    std::array<int, Cursor::ZERO.column> col_arr{};
    std::cout << "rows: " << row_arr.size() << "\ncols: " << col_arr.size();
    return 0;
}

Fully tested with GCC (ideone), clang (rextester), and MSVC++ 2017 (note that IntelliSense doesn't like it, but it compiles correctly!).

Answer 4

You can if you accept to have a function, not a variable

struct Cursor
{
    size_t row,column;

    static constexpr Cursor ZERO() { return Cursor{0,0}; }
};

Answer 5

A bit late, but here's a workaround if all you need is a constant:

struct Cursor
{
    struct CursorInit
    {
        int a, b;
        constexpr operator Cursor() const;
    };

    int row, column;
    static constexpr CursorInit ZERO {0,0};

    constexpr bool operator==(const Cursor& rhs) const
    {
        return row == rhs.row && column == rhs.column;
    }
};

inline constexpr Cursor::CursorInit::operator Cursor() const
{
    return Cursor{a, b};
}

static_assert(Cursor::ZERO == Cursor{0, 0});