4

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）？

4

1 回答 1

1

这很混乱，因为Derived当模板参数推导发生时并不完整Base。我认为显而易见的答案 - 通过Vector和Scalar明确 - 是不令人满意的。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> {};

为什么奇怪的限制不能使用typedef？我发现：

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。

于 2013-07-17T03:42:12.023 回答