在链表Ç复制节点 [英] Copying nodes in linked list C

问题描述

我试图复制在链表中的一个节点。我不知道如果我做正确。我试图使测试案例，但他们才没有成功。如果某些人能告诉我，如果我哪里错了，我做了什么吧，还什么来测试我的code的最佳方式。

 结构节点
{
        INT ID;
        字符的一面;
        INT数量;
        双价;
};结构onode
{
        结构节点*数据;
        结构onode *接下来的;
        结构onode * preV;
};结构onode * newNode（结构节点*数据）{
    结构节点* dataValue =（结构节点*）malloc的（的sizeof（结构节点））;
    结构onode *链表=（结构onode *）malloc的（的sizeof（结构onode））;    链表 - ＆GT;数据=（结构节点*）malloc的（的sizeof（数据）+1）;    如果（dataValue＆安培;＆安培;数据）
    {
        * dataValue = *的数据;
    }
}
 

我在我的code所做的更改，并添加更多的描述此功能需要什么。
一个改变：结构节点结构秩序

 结构秩序
{
        INT ID;
        字符的一面;
        INT数量;
        双价;
};结构onode
{
        为了结构*数据;
        结构onode *接下来的;
        结构onode * preV;
};
/ **
 *返回给定的顺序，功能填补了一个新的链接列表节点
 *分配一个新的订单，并复制存储在数据中的值，然后分配
 *链表节点。如果要实现此功能确保您
 *重复，作为原始数据可以由调用功能进行修改。
 * /结构onode * newNode（结构顺序*数据）
{
    为了结构* dataValue =（结构顺序*）malloc的（的sizeof（结构顺序））;
    结构onode *链表=（结构onode *）malloc的（的sizeof（结构onode））;    * dataValue = *的数据;    链表 - ＆GT;数据= dataValue;    linkedlist-＆GT;数据 - ＆GT; ID = dataValue-＆GT; ID;
    linkedlist-＆GT;数据 - ＆GT;价格= dataValue-＆GT;价格;
    linkedlist-＆GT;数据 - ＆GT;数量= dataValue-＆GT;数量;
    linkedlist-＆GT;数据 - ＆GT;侧= dataValue-＆GT;的一面;
    linkedlist-＆GT;下一步 - ＆GT; preV = NULL;    返回LinkedList的;}
 

解决方案

你的问题的关键是，你要创建两个新的节点对象 - 一个是 dataValue 键，一个是 linkedlist-＆GT;数据。然后，传入的数据复制到 dataValue 当你真的希望它被保存在 linkedlist-＆GT;数据

如果您替换

 链表 - ＆GT;数据=（结构节点*）malloc的（的sizeof（数据）+1）;
 

与

  linkedList-＆GT;数据= dataValue;
 

这应该让你在正确的方向前进。

I am trying to duplicate a node in linked list. I am not sure if I am doing it correctly. I tried making test cases but did they were not successful. If some one could tell me if where I went wrong and what I did right, also what is the best way to test my code.

struct node 
{
        int id;
        char side;
        int quantity;
        double price;
};

struct onode 
{
        struct node* data;
        struct onode* next;
        struct onode* prev;
};

struct onode* newNode (struct node* data)

{
    struct node* dataValue  = (struct node*) malloc(sizeof(struct node));
    struct onode* linkedlist = (struct onode*) malloc(sizeof(struct onode));

    linkedlist ->data = (struct node*)malloc(sizeof(data)+1);

    if(dataValue && data)
    {
        *dataValue = *data;
    }
}

I have made changes in my code and added more description on what this function wants. one change : struct node is struct order.

struct order 
{
        int id;
        char side;
        int quantity;
        double price;
};

struct onode 
{
        struct order* data;
        struct onode* next;
        struct onode* prev;
};


/**
 * Returns a new linked list node filled in with the given order, The function
 * allocates a new order and copy the values stored in data then allocate a 
 * linked list node. If you are implementing this function make sure that you
 * duplicate, as the original data may be modified by the calling function.
 */

struct onode* newNode (struct order* data)
{
    struct order* dataValue  = (struct order*) malloc(sizeof(struct order));
    struct onode* linkedlist = (struct onode*) malloc(sizeof(struct onode));

    *dataValue = *data;

    linkedlist ->data = dataValue;

    linkedlist->data->id = dataValue->id;
    linkedlist->data->price = dataValue->price;
    linkedlist->data->quantity = dataValue->quantity;
    linkedlist->data->side = dataValue->side;
    linkedlist->next->prev = NULL;

    return linkedlist;

}

解决方案

The crux of your problem is that you are creating two new node objects--one that is dataValue and one that is linkedlist->data. You then copy the passed-in data into dataValue when you really want it to be stored in linkedlist->data.

If you replace

linkedlist ->data = (struct node*)malloc(sizeof(data)+1);

with

linkedList->data = dataValue;

that should get you moving in the right direction.

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