雙鏈表C ++在第三個節點后崩潰

Question

因此，這是一個雙向鏈接的列表，應該包含名稱，地址和電話號碼並打印出來。 它適用於前三個節點，然后在第三個節點上輸入電話號碼后突然崩潰。 我相信這些指針有問題，但是我已經嘗試了所有我能想到的東西。

    #include <iostream>
    using namespace std;

class node
{
private:
    string elem;
    node* next;
    node* prev;
    string firstName;
    string lastName;
    string address;
    string phoneNumber;
    friend class linkedList;
};

//Linked list
class linkedList
{
public:
linkedList();
void addFrontNode(const string& e);
void addNode(const string& e);
void addNode2(node* nextloc, const string& e);
void addNode3(node* nextloc, const string& e);
void addNode4(node* nextloc, const string& e);
void print();
void search();
node* nextloc;
private:
node* head;
node* tail;


};

void linkedList::addFrontNode(const string &e)
{
node* v = new node;
string firstNameEntry;
string lastNameEntry;
string addressEntry;
string phoneNumberEntry;

cout << "Enter first name: ";
cin >> firstNameEntry;

cout << "Enter last name: ";
cin >> lastNameEntry;

cout << "Enter the address ";
cin.ignore();
getline(cin, addressEntry);

cout << "Enter a phone number ";
cin >> phoneNumberEntry;

v->elem = firstNameEntry;
v->lastName = lastNameEntry;
v->address = addressEntry;
v->phoneNumber = phoneNumberEntry;
v->next = head;
head = v;


}
void linkedList::addNode(const string &e)
{
node* v = new node;
string firstNameEntry;
string lastNameEntry;
string addressEntry;
string phoneNumberEntry;

cout << "Enter first name: ";
cin >> firstNameEntry;

cout << "Enter last name: ";
cin >> lastNameEntry;

cout << "Enter the address ";
cin.ignore();
getline(cin, addressEntry);

cout << "Enter a phone number ";
cin >> phoneNumberEntry;
v->elem = firstNameEntry;
v->lastName = lastNameEntry;
v->address = addressEntry;
v->phoneNumber = phoneNumberEntry;

v->next = tail;
tail = v;
tail->next = NULL;
}

void linkedList::addNode2(node* nextloc, const string &e)
{
node* v = new node;
string firstNameEntry;
string lastNameEntry;
string addressEntry;
string phoneNumberEntry;

cout << "Enter first name: ";
cin >> firstNameEntry;

cout << "Enter last name: ";
cin >> lastNameEntry;

cout << "Enter the address ";
cin.ignore();
getline(cin, addressEntry);

cout << "Enter a phone number ";
cin >> phoneNumberEntry;
v->elem = firstNameEntry;
v->lastName = lastNameEntry;
v->address = addressEntry;
v->phoneNumber = phoneNumberEntry;

nextloc = head -> next;
v->next = nextloc;

v->next = nextloc;
v->prev = nextloc->prev;
nextloc->prev = v;
}
void linkedList::addNode3(node* nextloc, const string &e)
{
node* v = new node;
string firstNameEntry;
string lastNameEntry;
string addressEntry;
string phoneNumberEntry;

cout << "Enter first name: ";
cin >> firstNameEntry;

cout << " Enter last name: ";
cin >> lastNameEntry;

cout << " Enter the address ";
cin.ignore();
getline(cin, addressEntry);

cout << " Enter a phone number ";
cin >> phoneNumberEntry;
v->elem = firstNameEntry;
v->lastName = lastNameEntry;
v->address = addressEntry;
v->phoneNumber = phoneNumberEntry;

v->next = nextloc;
v->prev = nextloc->prev;
nextloc->prev = v;
}

void linkedList::addNode4(node* nextloc, const string &e)
{
node* v = new node;
string firstNameEntry;
string lastNameEntry;
string addressEntry;
string phoneNumberEntry;

cout << "Enter first name: ";
cin >> firstNameEntry;

cout << " Enter last name: ";
cin >> lastNameEntry;

cout << " Enter the address ";
cin.ignore();
getline(cin, addressEntry);

cout << " Enter a phone number ";
cin >> phoneNumberEntry;
v->elem = firstNameEntry;
v->lastName = lastNameEntry;
v->address = addressEntry;
v->phoneNumber = phoneNumberEntry;

v->next = nextloc;
v->prev = nextloc->prev;
nextloc->prev->next = v;
nextloc->prev = v;
}
linkedList::linkedList() :head(NULL) {}

void linkedList::print()
{
node* v = new node;
v = head;
while (v != NULL)
{
    cout << v->elem << " ";
    cout << v->lastName << " ";
    cout << v->address << " ";
    cout << v->phoneNumber;
    v = v->next;
}
}
void linkedList::search()
{
node* v = new node;
v = tail;
string lastNameSearch;
cout << "Enter a last name to search ";
cin >> lastNameSearch;
while (v != NULL)
{
    if (v->lastName == lastNameSearch)
    {
        cout << v->elem;
        cout << v->address;
        cout << v->phoneNumber;

    }
    v = v->prev;
}
}
int main()
{
string node1;
string node2;
string node3;
string node31;
string node4;
string node5;

linkedList list;
list.addFrontNode(node1);
list.addNode(node2);
list.addNode2(list.nextloc, node3);
list.addNode3(list.nextloc, node4);
list.addNode4(list.nextloc, node5);
list.print();
return 0;
}

Answer 1

有幾個問題。

1. 如果使用addFrontNode()添加第一個節點，則必須設置tail 。 更改此：

    v->next = head; head = v; 

   對此：

    v->next = NULL; head = v; tail = v; 

2. 您的函數addNode()未正確添加到列表中，請嘗試調用print ，您將看到此函數未添加任何節點。 更改此：

    v->next = tail; tail = v; tail->next = NULL; 

   對此：

    tail->next = v; v->next = NULL; tail = v; 

3. 在main()只需使用addFrontNode()首先添加，然后使用addNode()添加所有其他。 之后，您的代碼按預期工作。

4. 不了解變量nextloc含義，可能是問題的根源。

總體建議：創建一個功能以添加節點

Answer 2

該代碼確實需要重寫，但是我不想重復第一個答案的建議。 但是，我認為問題在於墜機的原因。 這是因為在復制粘貼代碼時，您向addNode2添加了兩行：

void linkedList::addNode2(node* nextloc, const string &e)
{
...
nextloc = head -> next;
v->next = nextloc;
...
}

至少先注釋掉它們 ，它不會再崩潰（但這實際上並不能使它變得更好）。