使用概念Lite为具有成员函数模板的类型指定概念

Question

这是怎么做到的？

示例：假设我有较高的kinded类型HKT，其中有一个成员函数模板F:

template<class T>
struct HKT {
  template<class U> // this looks like e.g. rebind in std::allocators
  auto F(U) -> HKT<U>;
};

现在我想指定一个概念来约束这些更高级别的类型：

template <template <class> class HKT, class T>
concept HKTWithTemplateMemberFunctionF {
  return requires(HKT<T> h) { // HKT<T> is a type, h is an object
    // HKT<T> needs to have a member function template that 
    // returns HTK<U> where the type U is to be deduced and
    // it can be any type (it is unconstrained)
    template<class U>  // is there a syntax for this?
    h.F(std::declval<U>()) -> HKT<U>; 
  }
}

template <template <class> class HKT, class T, class U>
concept HKTWithTemplateMemberFunctionF {
  return requires(HKT<T> h) {
      h.F(std::declval<U>()) -> HKT<U>;
  }
}

Answer 1

长话短说，现在你（我？）必须提供特定的U:

template <template <class> class HKT, class T, class U = T>
concept HKTWithTemplateMemberFunctionF {
  return requires(HKT<T> h) { // HKT<T> is a type, h is an object
    h.F(std::declval<U>()) -> HKT<U>; 
  }
}

由于编译器无法为可能存在的所有类型u证明成员函数模板可以工作，也就是说，以下情况是无望的：

template <template <class> class HKT, class T>
concept HKTWithTemplateMemberFunctionF {
  return requires(HKT<T> h) {
    template<class U>  // for all those Us that haven't been written yet...
    h.F(std::declval<U>()) -> HKT<U>; 
  }
}

在一个假设的5分钟内指定概念的Lite实现中，我们能够对U进行一点点约束：

template <template <class> class HKT, class T, 
          InputIterator U = InputIterator()  /* imaginary syntax */ >
concept HKTWithTemplateMemberFunctionF {
  return requires(HKT<T> h) {
      h.F(std::declval<U>()) -> HKT<U>; // Is InputIterator enough to instantiate F?
  }
}