CODEWITHSUNDEEP
c++c++11constexpr
ndkrempel
ndkrempel

Reputation: 1936

static constexpr member of same type as class being defined

I would like a class C to have a static constexpr member of type C. Is this possible in C++11?

Attempt 1:

struct Foo {
    constexpr Foo() {}
    static constexpr Foo f = Foo();
};
constexpr Foo Foo::f;

g++ 4.7.0 says: 'invalid use of incomplete type' referring to the Foo() call.

Attempt 2:

struct Foo {
    constexpr Foo() {}
    static constexpr Foo f;
};
constexpr Foo Foo::f = Foo();

Now the problem is the lack of an initializer for the constexpr member f inside the class definition.

Attempt 3:

struct Foo {
    constexpr Foo() {}
    static const Foo f;
};
constexpr Foo Foo::f = Foo();

Now g++ complains about a redeclaration of Foo::f differing in constexpr.

Upvotes: 51

Views: 9106

Answers (5)

Daniel McLaury
Daniel McLaury

Reputation: 4273

Earlier I had the same problem and came across this decade-old question. I'm happy to report that in the intervening years a solution has appeared; we just need to do something like "attempt 3" above, but mark the definition of Foo::f as inline. Minimal example which compiles with g++ --std=c++17:

foo.hpp

#ifndef FOO_HPP
#define FOO_HPP

struct Foo
{
    constexpr Foo() {}
    static const Foo f;
};

inline constexpr Foo Foo::f = Foo();

#endif

foo.cpp

#include "foo.h"

main.cpp

#include "foo.h"

int main(int, char **) { return 0; }

Upvotes: 2

Ben
Ben

Reputation: 9703

If, like me, you are trying to make enum-like classes, the best I've figured out is to use CRTP to put the behavior in a base class and then a derived class is the "real" class that exists just to have the "enumerator"-like values as static constexpr inline Base members. This means that Foo::yes isn't of type Foo, but it acts like Foo and is implicitly convertible to Foo, so it seems pretty close. https://godbolt.org/z/rTEdKxE3h

template <class Derived>
class StrongBoolBase {
public:
    explicit constexpr StrongBoolBase() noexcept : m_val{false} {}
    explicit constexpr StrongBoolBase(bool val) noexcept : m_val{val} {}

    [[nodiscard]] constexpr explicit operator bool() const noexcept { return m_val; }

    [[nodiscard]] constexpr auto operator<=>(const StrongBoolBase&) const noexcept = default;
    [[nodiscard]] constexpr auto operator not() const noexcept {
        return StrongBoolBase{not this->asBool()};
    }
    [[nodiscard]] constexpr bool asBool() const noexcept {
        return m_val;
    }

private:
  bool m_val;
};

template <class Tag>
class StrongBool : public StrongBoolBase<StrongBool<Tag>> {
    using Base = StrongBoolBase<StrongBool<Tag>>;
public:
    //////// This is the interesting part: yes and no aren't StrongBool:
    inline static constexpr auto yes = Base{true};
    inline static constexpr auto no = Base{false};
    using Base::Base;
    constexpr StrongBool(Base b) noexcept : Base{b} {}    
};

The only breakdown is if you start to use decltype(Foo::yes) as though it's a Foo.

Upvotes: 0

xavlours
xavlours

Reputation: 753

An update on Richard Smith's answer, attempt 3 now compiles on both GCC 4.9 and 5.1, as well as clang 3.4.

struct Foo {
  std::size_t v;
  constexpr Foo() : v(){}
  static const Foo f;
};

constexpr const Foo Foo::f = Foo();

std::array<int, Foo::f.v> a;

However, when Foo is a class template, clang 3.4 fails, but GCC 4.9 and 5.1 still work ok:

template < class T >
struct Foo {
  T v;
  constexpr Foo() : v(){}
  static const Foo f;
};

template < class T >
constexpr const Foo<T> Foo<T>::f = Foo();

std::array<int, Foo<std::size_t>::f.v> a; // gcc ok, clang complains

Clang error :

error: non-type template argument is not a constant expression
std::array<int, Foo<std::size_t>::f.v> a;
                ^~~~~~~~~~~~~~~~~~~~~

Upvotes: 14

Richard Smith
Richard Smith

Reputation: 14158

I believe GCC is incorrect to reject your Attempt 3. There is no rule in the C++11 standard (or any of its accepted defect reports) which says that a redeclaration of a variable must be constexpr iff the prior declaration was. The closest the standard comes to that rule is in [dcl.constexpr](7.1.5)/1_:

If any declaration of a function or function template has constexpr specifier, then all its declarations shall contain the constexpr specifier.

Clang's implementation of constexpr accepts your Attempt 3.

Upvotes: 20

jogojapan
jogojapan

Reputation: 69957

If I interpret the Standard correctly, it isn't possible.

(§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. [...]

From the above (along with the fact that there is no separate statement about non-literal types in static data member declarations), I believe it follows that a static data member that is constexpr must be a literal type (as defined in §3.9/10), and it must have its definition included in the declaration. The latter condition could be satisfied by using the following code:

struct Foo {
  constexpr Foo() {}
  static constexpr Foo f {};
};

which is similar to your Attempt 1, but without the class-external definition.

However, since Foo is incomplete at the time of declaration/definition of the static member, the compiler can't check whether it is a literal type (as defined in §3.9/10), so it rejects the code.

Note that there is this post-C++-11 document (N3308) which discusses various problems of the current definition of constexpr in the Standard, and makes suggestions for amendments. Specifically, the "Proposed Wording" section suggests an amendment of §3.9/10 that implies the inclusion of incomplete types as one kind of literal type. If that amendment was to be accepted into a future version of the Standard, your problem would be solved.

Upvotes: 37

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