c – 带有基类的CRTP试图获取派生类成员的返回类型：无效使用不

发布时间：2020-12-16 07:05:11 所属栏目：百科来源：网络整理

导读：请考虑以下代码(仅用于示例目的)： #include iostream#include type_traits#include arraytemplate class Crtp,class Vector = typename std::decaydecltype(std::declvalCrtp().data())::type,class Scalar = typename std::decaydecltype(std::declvalCrtp

请考虑以下代码(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)？

解决方法

这非常混乱,因为当Base的模板参数推断发生时,Derived不完整.我假设显而易见的答案 – 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.

（编辑：李大同）

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