受到CppCon2016中Herb Sutter谈话的启发,可以在此链接中找到。
我决定使用智能指针实现视频中所示的双向链接列表。
以下实现几乎与remove()方法中的一行代码无关。
我调试了此代码,并且移除后,上一个节点未更新为null(应该是头节点)。
好像智能指针之间的所有权转移是错误的。以下是头文件和测试main()的代码:
LinkedList.h
#ifndef LINKEDLIST_H
#define LINKEDLIST_H
#include <iostream>
#include <memory>
#include <initializer_list>
namespace DLL {
template <typename T> class LinkedList{
private:
struct ListNode{
std::unique_ptr<ListNode> next; //2 uniq_ptr can't point to one another.
ListNode* prev = nullptr; //weakptr needs to be cast back to a shared_ptr to check its state.
T data{}; //Initialize empty;
ListNode(const T& element){
this->data = element;
}
};
public:
std::unique_ptr<ListNode> head;
ListNode* tail = nullptr;
LinkedList(){}
~LinkedList(){}
void append(const T& element){
ListNode* curr = nullptr;
if (head.get() == nullptr){ //If list is empty.
head = std::make_unique<ListNode>(element);
}
else if(head.get() -> next.get() == nullptr){ //If list has one element
head.get() -> next = std::make_unique<ListNode>(element);
curr = head.get() -> next.get(); //Sets raw pointer to the first element.
curr -> prev = head.get();
tail = curr;
}
else{
tail -> next = std::make_unique<ListNode>(element);
curr = tail -> next.get(); //Sets raw pointer to the last element.
curr -> prev = tail;
tail = curr;// The new last element is the tail.
}
}
int remove(const T& element){
ListNode* curr = nullptr;
if (head.get() == nullptr){ //If list is empty.
return -1; //Error: Can't remove from empty list.
}
//List has one or more elements.
curr = head.get();
while(curr != nullptr){
if(curr -> data == element){ //Found element
if(curr -> prev == nullptr){ //is head
//head.reset(head.get()->next.get()); Doesn't work
//Line below doesn't work too
head = std::move(curr->next); //Head now points to the next element
//New head's previous element doesn't point to nothing, as it should.
}
else if(curr -> next.get() == nullptr){ //is tail
tail = curr -> prev; //Reference the previous element
tail -> next.release(); //Release the old tail element
if(head.get() == tail){
tail = nullptr; //tail and head should not be the same.
} //List contains one element
}
else{//is intermediate
//The next node should point to the previous one
curr -> next -> prev = curr -> prev;
curr -> prev -> next = std::move(curr -> next);
//The prev node now points to the next one of current.
}
return 1; //Element found in list
}
curr = curr -> next.get(); //Traverse the next element
}
return 0; //Element not found in list
}
void print() {
ListNode* curr = head.get(); //Start from the start of the list.
std::cout << "[ ";
while (curr != nullptr) {
std::cout << curr -> data << " ";
curr = curr -> next.get();
}
std::cout << "]" << std::endl;
}
};
}
#endif
main.cpp
int main() { //Temporary Test Main will be split from the implementation file in the future
DLL::LinkedList <int> list; //Empty list
list.append(1);
list.append(4);
list.append(5);
list.append(6);
list.print();
list.remove(5);
list.remove(1); //When 1 is removed the 4 doesn't properly update as head, meaning the previous pointer of 4 is not null
list.remove(4);
list.remove(6);
list.print();
retunn 0;
}
对于此类问题,我感到很抱歉,我进行了很多搜索,但找不到类似的内容。我已经调试了好几天,但无法解决所有权问题。我尝试包含最少的代码,以重现该错误,如果标头是长代码段,对不起。
我用g ++:g++ -std=c++14 main.cpp -o out
和VS2015编译器编译。make_unique
呼叫需要C ++ 14标志
在您要检查迭代器的上一个指针为null的部分(基本上检查头)中,您具有以下行:
head = std::move(curr->next);
这会将标头指针移动为元素的下一个指针。但是,没有地方可以将新的头部指针的先前指针更新为null。因此该代码应为:
if(curr -> data == element){ //Found element
if(curr -> prev == nullptr){ //is head
head = std::move(curr->next); //Head now points to the next element
if (head)
head->prev = nullptr;
else
tail = nullptr;
}
}
由于您正在使用std::move
的对象将成为新的头部指针(这是正确的),因此基本上可以使该节点中包含的数据保持相同。在这种情况下,您需要明确-std::unique_ptr
包装器对其所拥有的对象的基础实现一无所知,因此prev
在这种情况下,它不知道如何更新指针。
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