C ++中的链接列表使用引用而不是指针 [英] Linked-list in C++ using references instead of pointers
问题描述
struct ListNode
{
int value;
ListNode * nextNode;
}
我的问题是....可以使用引用而不是指针?
struct ListNodeWithRefs
{
int value;
ListNodeWithRefs& nextNode;
}
我不知道它会提供任何性能收益,但...这个问题在编码时弹出,我的答案是否,但我可能会缺少某些东西。
原则上,没有什么可以阻止你使用引用和构造列表优化:
ListNodeWithRefs :: ListNodeWithRefs(ListNodeWithRefs& next):
nextNode
{}
但是有一个鸡蛋和鸡蛋问题,因为 next
还强制执行其下一个
元素在其创建时存在...等等...
注意:我认为我的问题也可以应用于将列表定义为:
struct ListNodeConst
{
int value;
const ListNode * nextNode;
}
通过sbi,它似乎工作: http://stackoverflow.com/a/3003607/1758762
//小心,未编译的代码!
template< typename T>
结构节点;
模板< typename T>
struct links {
node< T>&上一页
node& T&下一个;
link(node< T> * prv,node< T> * nxt); //省略
};
模板< typename T>
struct node {
T data;
链接< T> linked_nodes;
node(const T& d,node * prv,node * nxt); //省略
};
//技术上,这导致UB ...
模板< typename T>
void my_list< T> :: link_nodes(node< T> * prev,node< T> * next)
{
node< T> * prev_prev = prev.linked_nodes.prev;
node< T> * next_next = next.linked_nodes.next;
prev.linked_nodes。〜links< T>();
new(prev.linked_nodes)links< T>(prev_prev,next);
next.linked_nodes。〜links< T>();
new(next.linked_nodes)links< T>(next,next_next);
}
模板< typename T>
void my_list< T> :: insert(node< T> * at,const T& data)
{
node< T> * prev = at;
node< T> * next = at.linked_nodes.next;
node< T> * new_node =新节点< T>(data,prev,next);
link_nodes(prev,new_node);
link_nodes(new_node,next);
}
Suppose I want to create an unmodifiable linked-list (i.e. it can only be traversed, no nodes can be added or removed once it was initially created). This could be easily implemented by:
struct ListNode
{
int value;
ListNode* nextNode;
}
My question is .... Would it be possible to use references instead of pointers?
struct ListNodeWithRefs
{
int value;
ListNodeWithRefs &nextNode;
}
I am not sure it would provide any performance gain at all but ... this question popped up while coding and my answer so far is no but I could be missing something.
In principle, nothing prevents you from using references, and constructing list elments like this:
ListNodeWithRefs::ListNodeWithRefs(ListNodeWithRefs &next):
nextNode(next)
{}
But there is a chicken and egg problem because next
also enforces its next
element to exist at its creation and so on ...
Note: I think my question can also be applied to defining the list as:
struct ListNodeConst
{
int value;
const ListNode* nextNode;
}
Take a look at this example by sbi, it seems to work: http://stackoverflow.com/a/3003607/1758762
// Beware, un-compiled code ahead!
template< typename T >
struct node;
template< typename T >
struct links {
node<T>& prev;
node<T>& next;
link(node<T>* prv, node<T>* nxt); // omitted
};
template< typename T >
struct node {
T data;
links<T> linked_nodes;
node(const T& d, node* prv, node* nxt); // omitted
};
// technically, this causes UB...
template< typename T >
void my_list<T>::link_nodes(node<T>* prev, node<T>* next)
{
node<T>* prev_prev = prev.linked_nodes.prev;
node<T>* next_next = next.linked_nodes.next;
prev.linked_nodes.~links<T>();
new (prev.linked_nodes) links<T>(prev_prev, next);
next.linked_nodes.~links<T>();
new (next.linked_nodes) links<T>(next, next_next);
}
template< typename T >
void my_list<T>::insert(node<T>* at, const T& data)
{
node<T>* prev = at;
node<T>* next = at.linked_nodes.next;
node<T>* new_node = new node<T>(data, prev, next);
link_nodes(prev, new_node);
link_nodes(new_node, next);
}
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