双向链表混淆

Question

我创建了双向链表和 2 个函数。第一个函数从头到尾打印列表，第二个函数从头到尾打印列表。在第二个函数中，我将第一个 node->prev 设置为 NULL 但是

我不明白为什么第一个函数实际上有效，因为我从未将最后一个节点的下一个指针设置为 NULL，所以在我看来它应该创建一个无限循环。

代码如下：

#include <iostream>

using namespace std;

class LinkedList
{
    struct Node 
    {
        int data;
        Node * next;
        Node * prev;
    };

    public:
           LinkedList    ( void );
      void AddToList     ( int val );
      void FrontToBack   ( void ) const;
      void BackToFront   ( void ) const; 

    private:
        Node * head;
        Node * n;
        Node * tail;
};

LinkedList::LinkedList ( void )
{
    head = NULL;
    n = NULL;
    tail = NULL;
}

void LinkedList::AddToList ( int val )
{
    n = new Node ( );
    n -> data = val;
    if ( head == NULL )
    {
        n -> prev = NULL;
        head = n;
        tail = n;
    }
    else
    {
        n -> prev = tail;
        tail -> next = n;
        tail = n;
    }
}

void LinkedList::FrontToBack ( void ) const
{
    Node * tmp = head;
    int size = 0;

    cout << "Printing list from head to tail:" << endl; 

    while ( tmp != NULL )
    {
        if ( ! size )
        {
            cout << tmp -> data;
            tmp = tmp -> next;
        }
        else
        {
            cout << " " << tmp -> data;
            tmp = tmp -> next;
        }
        ++ size;
    }

    cout << endl;
}

void LinkedList::BackToFront ( void ) const
{
    Node * tmp = tail;
    int size = 0;

    cout << "Printing list from tail to head:" << endl;

    while ( tmp != NULL )
    {
        if ( ! size )
        {
            cout << tmp -> data;
            tmp = tmp -> prev;
        }
        else
        {
            cout << " " << tmp -> data;
            tmp = tmp -> prev;
        }
        ++ size;
    }

    cout << endl;
}



int main ( void )
{

    LinkedList list;

    list.AddToList( 1 );
    list.AddToList( 2 );
    list.AddToList( 3 );
    list.AddToList( 4 );
    list.AddToList( 5 );
    list.AddToList( 6 );

    list.FrontToBack( );
    list.BackToFront( );

    return 0;

}

Answer 1

I do not understand why the first function actually works because I have never set the last node's next pointer to NULL so in my opinion it should create an infinite loop.

struct Node 的 next 和 prev 指针从未被初始化，因此访问和取消引用它们是未定义的行为。这意味着您不能期望任何特定行为，例如 无限循环 。

为了确保定义的行为，只需默认初始化 Node 结构中的指针：

struct Node 
{
    int data;
    Node * next;
    Node * prev;

    Node() : next(nullptr), prev(nullptr) {} // <<<<<<<<<<<<<<<<<<<
};

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