Question

我正在尝试为模板化的双链表编写复制构造函数... 这是我到目前为止所做的：

// Copy Constructor
template <typename Item>
LinkedList<Item>::LinkedList(const LinkedList<Item> &s){
    // Create an iterator for the other list
    ListNode<Item> *node = s.head;
    // Create a reference to use as an iterator for the new list
    // Set it to the address of head of the new list
    ListNode<Item> **curr_node = &head;
    // Create a placeholder for the address of the previous node
    ListNode<Item> *prev_node = NULL;
    while(node != NULL){
            *curr_node = new ListNode<Item>;
            // If a prev_node address has been saved, initialize the new node's prev value
            // If a prev_node hasn't been saved yet, this means it is the head and prev should
            // be initialized to NULL 
            (*curr_node)->prev = prev_node;
            // Set the new node data fields to that of the node in the other list
            (*curr_node)->data = node->data;

            // -------- Set up for next iteration ------- 
            // Save the address of the current node to be used in the next node's prev_node 
            prev_node = *curr_node;
            // Set the curr_node pointer to the address of the next node in the new list 
            curr_node = &((*curr_node)->next);
            // Set the node pointer to the next node in other list
            node = node->next;
    }
    // Set the tail after all the nodes have been copied
    tail = prev_node;
}

当我在我的测试人员中调用此代码时：

LinkedList<int> ll;
ll.insert_back(5); 
ll.insert_back(10);
ll.insert_back(15);
LinkedList<int> ll1 = ll;
cout << "Printing contents of copied list: "; ll1.print();

我在valgrind中遇到此错误：

Contents of list to be copied: [5] -> [10] -> [15]
==4624== Conditional jump or move depends on uninitialised value(s)
==4624==    at 0x401077: LinkedList<int>::print() (LinkedList.cc:159)
==4624==    by 0x400D7B: main (tester.cpp:54)
==4624== 
==4624== Conditional jump or move depends on uninitialised value(s)
==4624==    at 0x40109E: LinkedList<int>::print() (LinkedList.cc:155)
==4624==    by 0x400D7B: main (tester.cpp:54)

我的print（）函数的第153,155,159行：

153: ListNode<Item> *end = head;
155: while(end != NULL){
159: if(end->next != NULL)

所以我得出的结论是，end永远不会被初始化，这意味着head是NULL或设置为某个垃圾值......任何人对此有什么想法？我哪里错了？

Answer 1

您似乎想要替换

行

curr_node = &((*curr_node)->next);

与

*curr_node = &(node->next);

因为curr_node之前在while循环中设置为new ListNode<Item>。

好奇，为什么curr_node是一个指向节点指针的指针？

复制双向链表的构造函数，初始化问题

1 个答案: