复制链表 C 中的节点 [英] Copying nodes in linked list C

问题描述

我正在尝试复制链表中的节点.我不确定我是否做对了.我尝试制作测试用例，但没有成功.如果有人能告诉我我哪里出错了，我做对了什么，还有什么是测试我的代码的最佳方法.

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;
    }
}

我对代码进行了更改，并添加了有关此功能所需内容的更多说明.一个变化:结构节点是结构顺序.

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;

}

推荐答案

问题的症结在于您正在创建两个新的 node 对象——一个是 dataValue> 和一个是 linkedlist->data.然后，当您确实希望将传入的数据存储在 linkedlist->data 中时，将其复制到 dataValue 中.

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.

如果更换

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

与

linkedList->data = dataValue;

这应该会让你朝着正确的方向前进.

that should get you moving in the right direction.

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