我正在使用 Visual Studio Community 2019 在 Windows 10 Pro（64 位）上處理反向鏈接串列。我想知道如何解決我遇到的錯誤，如下所示。我得到以下錯誤的while在成員函式回圈reverse()中class List
（*請假設該串列已經像連續整數一樣準備好了，例如 0,1,2,3,4,5）你能請任何人給我一些建議嗎？先感謝您。
拋出未處理的例外：讀取訪問沖突。 當前為 0xDDDDDDDD。
#include <iostream>
#include <string.h>
#include <vector>
using std::cout;
using std::endl;
template <class Data>
class Node
{
public:
Node* next;
Data data;
};
template <class Data>
class List
{
private:
Node<Data>* head;
Node<Data>* tail;
int count;
public:
//constructor
List()
{
count = 0;
head = nullptr;
tail = nullptr;
}
int size()
{
return count;
}
void msg_empty_list()
{
cout << "The list is not modified since it is empty." << endl;
}
int push_back(Data data)
{
Node<Data>* node = new Node<Data>;
node->data = data;
node->next = nullptr;
if (head == nullptr)
{
head = node;
tail = node;
}
else if (head != nullptr)
{
tail->next = node;
tail = tail->next;
}
count ;
return count;
}
int push_front(Data data)
{
Node<Data>* node = new Node<Data>;
node->data = data;
node->next = nullptr;
if (head == nullptr)
head = node;
else if (head != nullptr)
{
node->next = head;
head = node;
}
count ;
return count;
}
int pop_front(void)
{
if (head == nullptr)
return -1;
else if (head != nullptr)
{
Node<Data>* temp;
temp = head;
head = head->next;
delete temp;
count--;
return count;
}
}
int pop_back(void)
{
if (head == nullptr)
return -1;
else if (head != nullptr)
{
Node<Data>* temp = head;
while (temp->next != tail)
temp = temp->next;
delete tail;
tail = temp;
count--;
return count;
}
}
int remove_at(int index)
{
if (head == nullptr)
return -1;
else if (head != nullptr)
{
cout << "Specified index = " << index << endl;
if (index == 0)
{
pop_front();
}
else if (index == -1)
{
pop_back();
}
Node<Data>* temp = head;
Node<Data>* rmv;
int countIndex = 0;
while (countIndex < index - 1)
{
temp = temp->next;
countIndex ;
}
rmv = temp->next;
temp->next = temp->next->next;
delete rmv;
count--;
return count;
}
}
void reverse()
{
Node<Data>* temp = nullptr;
Node<Data>* prev = nullptr;
Node<Data>* current = head;
while (current != nullptr)
{
temp = current->next; // where I get error "Unhandled exception thrown: read access violation. **current** was 0xDDDDDDDD."
current->next = prev;
prev = current;
current = temp;
}
head = prev;
}
void print()
{
Node<Data>* temp = head;
if (head == nullptr)
return;
for (int i = 0; i < count - 1; i )
{
cout << temp->data << ", ";
temp = temp->next;
}
cout << temp->data;
}
~List()
{
Node<Data>* temp = head;
while (temp->next != nullptr)
{
temp = temp->next;
delete head;
head = temp;
}
}
};
int main()
{
Node<int> x;
Node<bool> y;
Node<char> n;
List<int> list;
//insert items into list
for (int i = 0; i < 10; i )
{
list.push_back(i);
}
cout << "Original list[size=" << list.size() << "]: ";
//print the list
list.print();
cout << endl;
// push a node to the beginning of the list
cout << endl << "==> Push a node to the head" << endl;
list.push_front(-1);
cout << "Modified list[size=" << list.size() << "]: ";
list.print();
cout << endl;
// pop the head of the list
cout << endl << "==> Pop a node from the head" << endl;
if (list.size() == 0)
list.msg_empty_list();
else
{
list.pop_front();
cout << "Modified list[size=" << list.size() << "]: ";
list.print();
cout << endl;
}
/*
// pop the tail of the list
cout << endl << "==> Pop a node from the tail" << endl;
if (list.size() == 0)
list.msg_empty_list();
else
{
list.pop_back();
cout << "Modified list[size=" << list.size() << "]: ";
list.print();
cout << endl;
}
*/
// delete the node at the specified index
cout << endl << "==> Delete the node at the specified index" << endl;
if (list.size() == 0)
list.msg_empty_list();
else
{
list.remove_at(5);
cout << "Modified list[size=" << list.size() << "]: ";
list.print();
cout << endl;
}
// reverse the list
cout << endl << "==> Reverse the list" << endl;
if (list.size() == 0)
list.msg_empty_list();
else
{
list.reverse();
cout << "Modified list[size=" << list.size() << "]: ";
list.print();
cout << endl;
}
return 0;
}
uj5u.com熱心網友回復：
我可以在您的代碼中發現一些錯誤，但讓我們專注于pop_back()函式：
int pop_back(void)
{
if (head == nullptr)
return -1;
else if (head != nullptr) // 1)
{
Node<Data>* temp = head;
while (temp->next != tail) // 2)
temp = temp->next;
delete tail;
tail = temp; // 3)
count--;
return count;
}
}
- 請不要這樣做：
if (a == nullptr) {...} else if (a != nullptr) {...}是多余的。離開第二個if。它讓讀者相信可能還有第三種情況。 - 幸運的是，這通常有效，但其他一些方法無法正確更新尾指標，因此這可能永遠不會正確。
pop_front當串列中只有一個元素時驗證您的方法。其他功能可能有類似的問題。 - 這是您正在觀察的實際問題。您沒有將
next新尾部元素的指標設定為空，而是指向現在已洗掉的尾部。tail->next = nullptr在此行之后插入。
轉載請註明出處，本文鏈接：https://www.uj5u.com/qiye/346264.html