处理链表数组

My approach:

我的方法：

An array of fixed-length (lets say 20) each element is pointer to the first node of a linked list. so i have 20 different linked list.

每个元素的固定长度（假设为20）数组是指向链表的第一个节点的指针。所以我有20个不同的链表。

This is the structure:

这是结构：

struct node{
       char data[16];
       struct node *next;
};

My declaration for that array

我对该数组的声明

struct node *nodesArr[20];

now to add a new node to one of the linked list, i do this:

现在要将一个新节点添加到其中一个链表，我这样做：

struct node *temp;

temp = nodesArr[i]; // i is declared and its less than 20
addNode(temp,word); // word is declared (char *word) and has a value ("hello")

The addNode function:

addNode函数：

void addNode(struct node *q, char *d){
    if(q == NULL)
        q = malloc(sizeof(struct node));
    else{
        while(q->next != NULL)
            q = q->next;

        q->next = malloc(sizeof(struct node));
        q = q->next;
    }

    q->data = d; // this must done using strncpy
    q->next = NULL; 
}

and to print data from the array of linked list, i do this:

并从链表列表中打印数据，我这样做：

void print(){
    int i;
    struct node *temp;

    for(i=0 ; i < 20; i++){
        temp = nodesArr[i];
        while(temp != NULL){
            printf("%s\n",temp->data);
            temp = temp->next;
        }
    }
}

now compiler gives no error, the program run and i pass the data to it, and when i call print it doesn't print any thing,,??

现在编译器没有错误，程序运行，我将数据传递给它，当我打电话打印它不打印任何东西，??

UPDATE::

更新::

after I edited the code (thx for you), i think the problem in the print function,, any idea ?

在我编辑代码（thx for you）之后，我认为打印功能中的问题，任何想法？

4 个解决方案

#1

The problem lies in addNode(). When the list is empty you do:

问题在于addNode（）。当列表为空时，您执行以下操作：

q = malloc(sizeof(struct node));

but the scope of q is limited to addNode(). You should have declared addNode() as

但q的范围仅限于addNode（）。您应该将addNode（）声明为

void addNode(struct node **q, char *d)

and adjust your code accordingly:

并相应地调整您的代码：

*q = malloc(sizeof(struct node));

and so on...

等等...

#2

When you pass struct node *q to addNode you are giving it an address for an element in your array. If you use malloc inside, then you are overwriting this variable q, which is local to the function and now points to something different, but you haven't changed your original array. Try using a pointer to pointer to node (struct node **q).

当您将struct node * q传递给addNode时，您将为其提供数组中元素的地址。如果你在里面使用malloc，那么你将覆盖这个变量q，它是函数的本地变量，现在指向不同的变量，但你还没有改变原始数组。尝试使用指向节点的指针（struct node ** q）。

#3

void addNode(struct node *q, char *d){
    if(q == NULL)
        q = malloc(sizeof(struct node));

Here's the problem.

这是问题所在。

The new value of q doesn't ever get out of the function, so your array of linked lists never gets updated.

q的新值永远不会离开函数，因此链接列表的数组永远不会更新。

Normally the solution here is to use a double-pointer:

通常这里的解决方案是使用双指针：

void addNode(struct node **q, char *d){
    if(*q == NULL)
        *q = malloc(sizeof(struct node));

And call it like so:

并称之为：

addNode(&nodesArr[i],word);

Then, if you malloc a new node, the value in the array will be set to point to the new node.

然后，如果malloc新节点，则数组中的值将设置为指向新节点。

#4

-2

struct node
{

  int actual, estimated;

  char c;

  struct node *next;

} *head[4], *var[4], *trav[4];


void
insert_at_end (char c, int value, int value1)
{

  struct node *temp;

  temp = head[i];

  var[i] = (struct node *) malloc (sizeof (struct node));

  var[i]->actual = value;

  //var1=(struct node *)malloc(sizeof(struct node));

  var[i]->estimated = value1;

  var[i]->c = c;

  //printf("%d",var->estimated);

  if (head[i] == NULL)

    {

      head[i] = var[i];

      head[i]->next = NULL;

    }

  else

    {

      while (temp->next != NULL)

    {

      temp = temp->next;

    }

      var[i]->next = NULL;

      temp->next = var[i];

    }

}

#1