Question

考虑以下代码（仅用于示例目的）：

#include <iostream>
#include <type_traits>
#include <array>

template <
class Crtp, 
class Vector = typename std::decay<decltype(std::declval<Crtp>().data())>::type, 
class Scalar = typename std::decay<decltype(std::declval<Crtp>().data(0))>::type
>
struct Base
{;};

template <
class Vector = std::array<double, 3>, 
class Scalar = typename std::decay<decltype(std::declval<Vector>()[0])>::type
>
struct Derived
: public Base<Derived<Vector, Scalar>>
{
    Vector _data;
    inline Vector& data() {return _data;}
    inline const Vector& data() const {return _data;}
    inline Scalar& data(const unsigned int i) {return _data[i];}
    inline const Scalar& data(const unsigned int i) const {return _data[i];}
};

int main()
{
    Derived<> d;
    return 0;
}

它返回以下错误：

main.cpp: In instantiation of 'struct Derived<>':
main.cpp:28:14: required from here
main.cpp:16:8: error: invalid use of incomplete type 'struct Derived<>'
main.cpp:16:8: error: declaration of 'struct Derived<>'

有没有办法解决这个问题（不使用typedef，只使用模板）？

Answer 1

这非常混乱，因为当Derived发生模板参数推断时Base未完成。我认为明显的答案 - 明确地传递Vector和Scalar - 是不能令人满意的。怎么样：

template <template <class, class> class Derived,
          class Vector, class Scalar>
struct Base {};

template <class Vector, class Scalar>
struct Derived : Base<Derived, Vector, Scalar> {};

为什么不使用typedef的奇怪限制？我找到了：

template <class Vector>
using ScalarTypeOf =
  typename std::decay<decltype(std::declval<Vector>()[0])>::type;

template <class Crtp>
using VectorTypeOf =
  typename std::decay<decltype(std::declval<Crtp>().data())>::type;

template <class Crtp>
struct Base {
  using Vector = VectorTypeOf<Crtp>;
  using Scalar = ScalarTypeOf<Vector>;
};

template <class Vector>
struct Derived : public Base<Derived<Vector>> {
  using Scalar = ScalarTypeOf<Vector>;
};

更具可读性。

CRTP与基类试图获取派生类成员的返回类型：无效使用不完整类型

1 个答案: