[英]Implementation of a Doubly Linked List
我正在嘗試實現一個鏈表,我完全迷路了。 我到處都有斷點,特別是使用擦除方法。 每當我更改擦除方法時,都會不可避免地出現一些錯誤。 我遇到了指針錯誤,析構函數的問題,這些問題僅在調用擦除方法時才會發生,等等。
這是我到目前為止的內容:
頭文件:
#pragma once
class IntList {
private:
class IntNode {
public:
IntNode(int v, IntNode *pr, IntNode *nx);
~IntNode();
IntNode* previous;
IntNode* next;
class iterator {
public:
iterator(IntNode* t);
int& operator*();
iterator& operator++();
iterator& operator--();
bool operator!=(iterator other)const;
private:
IntNode* target;
};
private:
int value;
};
IntNode* head;
IntNode* tail;
int count;
public:
IntList();
~IntList();
void push_back(int v);
void pop_back();
int size() const { return count; }
typedef IntNode::iterator iterator;
iterator begin();
iterator end();
//unsigned int size() const;
void push_front(int value);
bool empty() const;
int& front();
int& back();
void clear();
iterator erase(iterator position);
};
執行:
#include "IntList.h"
#include <stdexcept>
IntList::IntList() : head{ nullptr }, tail{ nullptr }, count{ 0 }
{}
IntList::~IntList() {
while (head) {
head = head->next;
delete head;
}
}
void IntList::push_back(int v) {
tail = new IntNode{ v, tail, nullptr };
if (!head) { head = tail; }
count += 1;
}
void IntList::pop_back() {
tail = tail->previous;
delete tail->next;
count -= 1;
}
IntList::iterator IntList::begin()
{
return iterator{ head };
}
IntList::iterator IntList::end() {
return iterator{ nullptr };
}
void IntList::push_front(int value) {
head = new IntNode{ value, nullptr, head };
if (!tail) { tail = head; }
count += 1;
}
bool IntList::empty() const{
return (count==0);
}
int& IntList::front() {
return *begin();
}
int& IntList::back() {
return *begin();
}
void IntList::clear() {
head = nullptr;
tail = nullptr;
count = 0;
}
IntList::iterator IntList::erase(iterator position) {
int midpointL = 0;
for (iterator index = begin(); index != position; ++index) {
midpointL++;
}
if (midpointL == 0) {
head = head->next;
}
else if (midpointL == count) {
tail = tail->previous;
}
else {
// Move head to get a reference to the component that needs to be deleted
for (int i = 0; i < midpointL; i++) {
head = head->next;
}
// Change the previous and next pointers to point to each other
(head->previous)->next = (head->next);
(head->next)->previous = (head->previous);
for (int i = midpointL-1; i > 0; i++) {
head = head->previous;
}
}
count-=1;
return position;
}
IntList::IntNode::IntNode(int v, IntNode * pr, IntNode * nx)
: previous{ pr }, next{ nx }, value{ v }
{
if (previous) { previous->next = this; }
if (next) { next->previous = this; }
}
IntList::IntNode::~IntNode() {
if (previous) previous->next = next;
if (next) next->previous = previous;
}
IntList::IntNode::iterator::iterator(IntNode* t)
: target{ t }
{}
int& IntList::IntNode::iterator::operator*() {
if (!target) { throw std::runtime_error{ "Deferenced sentinel iterator." }; }
return target->value;
}
IntList::IntNode::iterator& IntList::IntNode::iterator::operator++()
{
if (target) { target = target->next; }
return *this;
}
IntList::IntNode::iterator& IntList::IntNode::iterator::operator--()
{
if (target) { target = target->previous; }
return *this;
}
bool IntList::IntNode::iterator::operator!=(iterator other)const
{
return (!(target == other.target));
}
有人可以幫我指出正確的方向嗎?
謝謝!
讓我們在這里進行一下快速回顧:
IntList::~IntList() {
while (head) {
head = head->next;
delete head;
}
}
您應該改為:
IntList::~IntList() {
while (head) {
IntNode* newHead = head->next;
delete head;
head = newHead;
}
}
當您刪除“下一個”對象,然后嘗試在下一次迭代中訪問它時。
void IntList::pop_back() {
tail = tail->previous;
delete tail->next;
count -= 1;
}
在這里,您無需檢查tail是否為null或是否指向head ..(什么是空條件?),也許count!=0
? 萬一您可以刪除不存在的下一節點
IntList::iterator IntList::end() {
return iterator{ nullptr };
}
.. end
是否為空? ebd應該是你的尾巴...
int& IntList::back() {
return *begin();
}
那是開始..不回來。
void IntList::clear() {
head = nullptr;
tail = nullptr;
count = 0;
}
一個清除對象應釋放列表中的所有對象。 您正在此處生成垃圾(泄漏)。
我在這里停下來,對此感到抱歉,這只是喝咖啡休息時間。 但是您應該仔細看一下:*空指針的使用*不需要時刪除節點列表項*注意不要使用無效的指針(例如head->previous->next
我見過的某個地方)
您必須自下而上檢查代碼。 希望這些第一提示可以幫助您完成學習過程。
玩得開心,Ste
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