复制链表C中的节点

我正在尝试复制链表中的节点。我不确定我是否做得正确。我尝试制作测试用例，但没有成功。如果有人可以告诉我哪里出了问题以及我做对了什么，以及测试我的代码的最佳方法是什么。

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，并添加了有关此函数所需内容的更多描述。 一项更改：结构节点是结构顺序。

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;

}

问题的关键是你正在创建两个新的node对象——一个是dataValue和一个是linkedlist->data。然后将传入的数据复制到dataValue当您确实希望将其存储在linkedlist->data.

如果你更换

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

with

linkedList->data = dataValue;

这应该会让你朝着正确的方向前进。