c++ - 不调用部分专用模板类(用于容器类类型)

Question

我仍在处理几个小时前发布的这个问题:[ How to overload/specialize template class function to handle arithmetic types and a container-class我试图实现这个解决方案。它编译但对象是使用 DerivedClass-Constructor 而不是部分专用模板类 DerivedClass< Eigen::ArrayBase > 创建的你知道我犯了(或一些)错误吗？

template <typename T> class BaseClass
{
protected:
  T mem;

public:
  BaseClass(T arg) : mem(arg){};
};

template <typename T> class DerivedClass : public BaseClass<T>
{
public:
  DerivedClass(T arg): BaseClass<T>(arg){};
};

template <typename T>
class DerivedClass<Eigen::ArrayBase<T> >
    : public DerivedClass<Eigen::ArrayBase<T> >
{
public: 
  DerivedClass(Eigen::ArrayBase<T> arg):BaseClass<Eigen::ArrayBase<T> >(arg){};
};

int main
{
...
  Eigen::Array3d arg = Array3d::Random(3);
  DerivedClass<Eigen::Array3d> o(arg);
....
}

Answer 1

template<template<class...>class Z>
struct template_instance_test {
  static std::false_type test(...);
  template<class...Ts>
  static std::true_type test( Z<Ts...> const* );
  template<class X>
  using tester = decltype(test( std::declval<X*>() ) );
};
template<template<class...>class Z, class T>
using is_derived_from_template = typename template_instance_test<Z>::template tester<T>;

我们现在可以询问某物是否是特定模板的实例，或者是否派生自特定模板的实例。

template<class X>
struct Base {};
template<class X>
struct Derived:Base<X> {};

template<class T>
struct Storage {
  T data;
};
template<class T, class=void>
struct Instance:Storage<T> {
  enum {is_special = false};
};
template<class T>
struct Instance<T, std::enable_if_t< is_derived_from_template<Base, T>{} > >:
  Storage<T> {
  enum { is_special = true };
};

int main() {
    Instance<int> i; (void)i;
    static_assert(!Instance<int>::is_special);
    Instance<Derived<int>> j; (void)j;
    static_assert(is_derived_from_template<Base, Base<int>>{});
    static_assert(is_derived_from_template<Base, Derived<int>>{});
    static_assert(Instance<Derived<int>>::is_special);
}

我们完成了。 Live example .