用智能指针实现一个简单的单向链表
我正在尝试使用智能指针实现一个简单的单链表，这是我到目前为止所拥有的，我选择使用 C++ 的 shared_ptr，但我读到 unique_ptr 更适合这种情况，但是，我不真的不知道您将如何遍历列表(即 currentNode = currentNode->next)以到达列表的末尾，以便使用 unique_ptr 插入一个元素.这是我到目前为止的代码:
Hi I'm trying to implement a simple singly linked list using smart pointers, here is what I have so far, I opted with using C++'s shared_ptr but I read that a unique_ptr would be more appropriate for this case but, I don't really know how you would iterate over the list (i.e currentNode = currentNode->next) to get to the end of the list in order to insert an element using a unique_ptr. Here is the code I have so far:
template <typename T>
class LinkedList;
template <typename T>
class ListNode
{
public:
ListNode() : _data(T()) {}
explicit ListNode(const T& value) : _data(value) {}
friend class LinkedList < T > ;
private:
T _data;
shared_ptr<ListNode<T>> _next;
};
template <typename T>
class LinkedList
{
public:
void push_back(const T& value)
{
if (_root)
{
shared_ptr<ListNode<T>> currentNode(_root);
while (currentNode->_next != nullptr)
{
currentNode = currentNode->_next;
}
currentNode->_next = make_shared<ListNode<T>>(value);
}
else
{
// If the list is completely empty,
// construct a new root (first element)
_root = make_shared<ListNode<T>>(value);
}
}
void print() const
{
shared_ptr<ListNode<T>> currentNode(_root);
while (currentNode != nullptr)
{
cout << currentNode->_data << " ";
currentNode = currentNode->_next;
}
cout << endl;
}
private:
shared_ptr<ListNode<T>> _root;
};
如果使用 unique_ptrs 是该程序的更好方法，您能否说明我将如何解决迭代问题?由于无法分配unique_ptrs，我将如何做代码块:
If using unique_ptrs are the better way to go for this program, could you illustrate how I would get past the iterating problem? Since unique_ptrs can't be assigned, how would I do the code block:
shared_ptr<ListNode<T>> currentNode(_root);
while (currentNode->_next != nullptr)
{
currentNode = currentNode->_next;
}
currentNode->_next = make_shared<ListNode<T>>(value);
使用 unique_ptrs 而不是 shared_ptrs?谢谢！
using unique_ptrs instead of shared_ptrs? Thanks!
推荐答案
您使用 std::unique_ptr 的循环可能如下所示:
Your loop with std::unique_ptr may look like:
// Iteration doesn't own resource, so no unique_ptr here.
ListNode<T>* currentNode(_root.get());
while (currentNode->_next != nullptr)
{
currentNode = currentNode->_next.get();
}
currentNode->_next = make_unique<ListNode<T>>(value);
相关文章