【发布时间】:2023-03-17 05:43:02
【问题描述】:
我正在编写自己的容器类,它也提供迭代器。这些迭代器可以被取消引用并显示原始容器的子范围,再次可以获得迭代器。
目前,我有一个模板迭代器类(使用boost::iterator_facade),如果L!=0 取消引用Collection(“范围”),如果L==0 取消引用T&(存储元素)。是否可以将两者结合在一个类中,从而需要更少的重复代码?
template<typename T, int L>
class CollectionIter : public boost::iterator_facade<
CollectionIter<T,L>, // type it selfe
Collection<T,L-1>, // value type
boost::random_access_traversal_tag,
Collection<T,L-1> > // deref. type
{
public:
CollectionIter(T* ptr, const std::vector<int>& collectionSize_)
: pointer(ptr), collectionSize(collectionSize_) { }
T* element() { return pointer; }
private:
friend class boost::iterator_core_access;
bool equal(const CollectionIter<T,L>& other) const { return pointer==other.pointer; }
auto dereference() const { return Collection<T,L-1>(pointer, collectionSize); }
void increment() { pointer = pointer + stepsize(); }
void decrement() { pointer = pointer - stepsize(); }
void advance(size_t i) { pointer = pointer + i*stepsize(); }
auto distance_to(const CollectionIter<T,L>& other) { return (other.pointer - pointer)/stepsize(); }
int stepsize() { return collectionSize.at(L); }
T* pointer;
const std::vector<int>& collectionSize;
};
/* Groundlevel Collection: deref returns T& */
template<typename T>
class CollectionIter<T,0> : public boost::iterator_facade<
CollectionIter<T,0>,
T,
boost::random_access_traversal_tag >
{
public:
CollectionIter(T* ptr, const std::vector<int>& collectionSize_)
: pointer(ptr), collectionSize(collectionSize_) { assert(stepsize()==1); }
T* element() { return pointer; }
private:
friend class boost::iterator_core_access;
bool equal(const CollectionIter<T,0>& other) const { return pointer==other.pointer; }
T& dereference() const { return *pointer; }
void increment() { pointer = pointer + stepsize(); }
void decrement() { pointer = pointer - stepsize(); }
void advance(size_t i) { pointer = pointer + i*stepsize(); }
auto distance_to(const CollectionIter<T,0>& other) { return (other.pointer - pointer)/stepsize(); }
int stepsize() { return collectionSize.at(0); }
T* pointer;
const std::vector<int>& collectionSize;
};
【问题讨论】:
-
如果不深入了解您的代码,您似乎想在代码中的某处应用
std::conditional以避免重复并仍然获得正确的继承。
标签: c++ templates c++14 template-meta-programming template-specialization