双链表核心转储

Question

在此先感谢。

我正在做一个双向链表。

一切工作正常，但我意识到，当我在中间的某个地方添加了一个新的类节点时，左指针仍然指向先前的节点（现在有两个空格）。

所以我增加了一个新的节点指针上线46

然后在第51行，我告诉那个节点现在指向新节点。

所以：

• 首先，我在太空中有了新节点临时关闭

• 然后我让指针temp2遍历列表

• 最后我告诉temp3指向节点在temp2的节点

功能运行后，顺序应该是temp2->temp->temp3

我主要：后我加线51，我的程序核心转储（分段错误），并关闭了。

我该如何解决这个问题？只有当我添加一些不在头指针的位置时才会发生。

void add(node *&head, node *&tail, node *&current) 
{ 
    node *temp = new node; //creates a pointer pointing to a new class node 
    cin >> temp->letter; // user input 

    current = head; // creates a pointer to point at the first node 
    while (current != NULL) // while list isn't empty 
    { 
     if (current->letter == temp->letter) 
     { // letter already exists 
      cout << "DUPLICATE: " << temp->letter << endl << endl; 
      return; 
     } 
     else 
     { // loop through list moving tail pointer to the end while checking for duplicates 
      tail = current; 
      current = current->right_link; 
     } 
    } 

    current = temp; // current = new added node 

    if (isEmpty(head)) 
    { // if first node 
     temp->left_link = NULL; 
     temp->right_link = NULL; 
     head = temp; // head and 
     tail = temp; // tail both point to first and only node. 
    } 
    else 
    { // if new letter value is less than head value 
     if(temp->letter < head->letter) 
     { 
      temp->right_link = head; // node points (right) to head 
      temp->left_link = NULL; // left most node point to nothing. 
      head->left_link = temp; // head (currently the second node) points (left) to first node 
      head = temp; // head pointer moves to the first position 
     } 
     else 
     { // if new node goes anywhere other than head 
      node *temp2 = head; // new node to cycle through list 
      while(temp2->right_link != NULL && temp2->right_link->letter < temp->letter) 
      { // if temp2 points to a node and that node's value is less than temp node value 
       temp2 = temp2->right_link; 
      } 
      node *temp3 = temp2->right_link; 
      temp->right_link = temp2->right_link; // when temp2 stops looping, temp will point to 
                // the same node as temp2. 
      temp2->right_link = temp; // temp2's current node will point to temp, causing temp 
           // to be added into the list (after temp2) 
      temp3->left_link = temp; // point the node (after the newly inserted node) left to new node 
      temp->left_link = temp2; // connects the left pointer between temp and temp2 
      if(temp->right_link == NULL) 
       tail = temp; 
     } 
    } 
    cout << "ADDED : " << temp->letter << endl << endl; 
} 

Answer 1

如果temp2->right_link == NULL

46  node *temp3 = temp2->right_link; 

是一个NULL指针，所以你不能

51  temp3->left_link = temp; 

，如果你使用的调试器，它应该是很明显。