2

我有一个链接列表，其中包含另一个链接列表，我想在其中集成数据，但我做不到。

这是我的代码：

结构声明：

typedef struct BigStructure {
    UINT x;
    UINT y;

    struct SmallStructure* smallStructure;

    struct BigStructure* next;
} BigStructure;

typedef struct SmallStructure {
    UINT x;
    UINT y;

    struct SmallStructure* next;
} SmallStructure;

结构操作函数：

BigStructure* addLinkedListElement(BigStructure* linkedList)
{
    if(linkedList-> next == NULL)
    {
        return NULL;
    }

    BigStructure* newLinkedList = malloc(sizeof(linkedList));
    newLinkedList->next = linkedList;
    return newLinkedList;
}

BigStructure* removeLinkedListElement(BigStructure* linkedList)
{
    //If the list is empty, we return NULL
    if(linkedList == NULL)
        return NULL;

    //If the list contains one element
    if(linkedList->next == NULL)
    {
        free(linkedList);
        return NULL;
    }

    //if the list contains at least 2 elements
    BigStructure* tmp = linkedList;
    BigStructure* ptmp = linkedList;

    /* Tant qu'on n'est pas au dernier élément */
    while(tmp->next != NULL)
    {
        //ptmp stores the address of tmp
        ptmp = tmp;
        //We move tmp (but pmpt keeps the old value of tmp)
        tmp = tmp->next;
    }

    ptmp->next = NULL;
    free(tmp);
    return linkedList;
}

BigStructure* getLinkedListElement(BigStructure* linkedList, int id)
{
    int i = 0;

    for(i=0; i<id && linkedList != NULL; i++)
    {
        linkedList = linkedList->next;
    }

    if(linkedList == NULL)
    {
        return NULL;
    }
    else
    {
        return linkedList;
    }
}

我尝试使用上面的代码来访问 SmallStructure 变量，但我得到了一个很大的数字（看起来像一个地址）：

BigStructure* bigStructure = NULL;

void addBigStructure(UINT x, UINT y) {

        if(bigStructureNb == 1)
        {
            bigStructure->x = x;
            bigStructure->y = y;
        }
        else
        {
            BigStructure* newBigStructure;
            newBigStructure = (BigStructure*)addLinkedListElement((BigStructure*)&bigStructure);
            newBigStructure->x = x;
            newBigStructure->y = y;
        }
}

void addSmallStucture(UINT x, UINT y) {

    if(smallStructuresNb == 1)
    {
        bigStructure->startTrigger = malloc(sizeof(BigStructure*));
        bigStructure->startTrigger->x = x;
        bigStructure->startTrigger->y = y;
    }
    else
    {
        BigStructure* tmpBigStructure = NULL;
        tmpBigStructure = (BigStructure*)getLinkedListElement(&bigStructure, rowID); //Table row ID
        g_print("%d", tmpBigStructure->id); //Here I get a false value !!!!
        //Here I want to set the value of the tmpBigStructure->smallStructure->x/y
}
}
4

2 回答 2

1

在我看来，问题出在 getLinkedListElement() 上。这是一些代码建议：

  BigStructure* getLinkedListElement(BigStructure** linkedList, int id)
    {
        int i;
        if( linkedList == NULL || *linkedList == NULL)
        return NULL ; 

     //We cannot update HEAD(linkedList), therfore using local pointer.
        BigStructure* linkWalk = * linkedList;

    /*I am asuming ids are mapped to linked list nodes as below.
     id 0 -> first node
     id 1 -> second node
     ......
     id n -> n-1 node
    */

//starting from second node since linkWalk is already pointing to first above.
        for(i=1; i<id && linkWalk != NULL; i++)  
                linkWalk = linkWalk->next;

   // At this point , either id = 0 OR id = i OR the link has been traversed.

       return linkWalk ;

    }

最后，在调用 g_print("%d", tmpBigStructure->id) 之前，请检查 tmpBigStructure != NULL。

于 2013-06-01T12:32:06.133 回答
0

在我看来，全局：“BigStructure* bigStructure = NULL”是你的头指针，你传递了它的地址。你可以在头上添加节点。因此，你不需要遍历列表来添加链接节点。

建议：您正在传递 bigStructure 的地址（即双指针）并在结构操作函数中将其作为单指针进行操作。这需要在您的所有功能中进行更改。

例如，您添加节点的函数可能类似于（假设 linkedList 是 HEAD）：

BigStructure* addLinkedListElement(BigStructure** linkedList)
{
    BigStructure* newLinkedList = malloc(sizeof(BigStructure));
    if (newLinkedList == NULL)
      return NULL ;    // Just to handle malloc failure

    newLinkedList->next = *linkedList;
    *linkedList = newLinkedList;
    return newLinkedList;
}
于 2013-06-01T07:49:18.127 回答