用C链接列表实现 [英] Linked List Implementation in C

问题描述

我是新来的链表，我想实现在C链表。
下面我code： -

I am new to Linked LIsts and I am trying to implement a Linked List in C . Below in my code :-

#include<stdio.h>
#include<stdlib.h>

struct node {
    int data;
    struct node *next;
};
void insert (struct node *head, int data);
void  print  (struct node *head);
int main()
{
    struct node *head ;
    head= NULL;
    printf("new\n");
    insert(head,5);
    printf("%d\n",head);
    insert(head,4);
    insert(head,6);
    print(head);
    print(head);
    print(head);


} 
void  insert(struct node *head,int data){

    printf("%d\n",head);
    if(head == NULL){
        head =malloc(sizeof(struct node));
        head->next = NULL;
        head->data = data;

    }
    else {
        printf("entered else\n");
        struct node *tmp = head;
        if(tmp->next!=NULL){
            tmp = tmp->next;
        }
        tmp->next  = malloc(sizeof(struct node));
        tmp->next->next = NULL;
        tmp->next->data = data;

    }

}


void print (struct node *head) {
    printf("%d\n",head);
    struct node *tmp = head;
    if (head == NULL) {
        printf("entered null\n");
        return;
    }
    while (tmp != NULL) {
        if (tmp->next == NULL) {
            printf("%0d", tmp->data);
        } else {
            printf("%0d -> ", tmp->data);
        }
        tmp = tmp->next;
    }
    printf("\n");
}

当我运行这个code输出： -

When I run this code the output is :-

new
0
0
0
0
0
entered null
0
entered null
0
entered null

头总是空并且它不更新空。它不进入中插入其他循环。
谁能帮我解决这个问题吧。指出我做的错误。
谢谢

The head is always null and it doesnt update the null . It doesnt enter into the else loop in insert . Can anyone help me fix this please . Point out the mistake I am doing . thanks

推荐答案

有可能在你的code其他错误，但一个大问题是，你正试图设置头节点插入，但只影响传入的指针的本地副本，所以它在主叫方没有任何影响：

There may be other errors in your code, but one big issue is that you are attempting to set a head node in insert, but that only affects a local copy of the pointer passed in, so it has no effect in the caller side:

void  insert(struct node *head,int data){
  ....
  head = malloc(sizeof(struct node)); // head is local, caller won't see this

您还需要确保当你通过这不是一个节点 NULL ，你居然attatch新节点的头部。
您可以通过传递一个指针的指针，或返回集指针解决的第一个问题。例如，

You also need to ensure that when you pass a node that is not NULL, you actually attatch the new node to the head. You can fix the first problem by passing a pointer to a pointer, or by returning the set pointer. For example,

void insert(struct node **head, int data) {
  if(*head == NULL) {
    // create the head node
    ...
    *head = malloc(sizeof(struct node)); 
    ....
  else {
    // create a new node and attach it to the head
    struct node* tmp = malloc(sizeof(struct node));
    ....
    (*head)->next = tmp;
  }
}

然后，在主，你需要一个指针传递给头指针，即使用地址运算符＆安培; ：

Then, in main, you need to pass a pointer to the head pointer, i.e. use the address-of operator &:

struct node *head = NULL;
insert(&head, 5);

---

注意的部分问题是该函数试图做太多。这就是所谓的插入，但它试图创建一个新的节点，如果传入的指针 NULL 。这将是更好的分离这些职责：

---

Note part of the problem is that the function is trying to do too much. It is called insert, but it attempts to create a new node if the pointer passed in is NULL. It would be better to separate these responsibilities:

// allocate a node and set its data field
struct node* create_node(int  data)
{
  struct node* n = malloc(sizeof(struct node));
  n->next = NULL;
  n->data = data;
  return n;
}

// create a node and add to end node.
// return the new end node.
// end has to point to a valid node object.
struct node* append_node(struct node* tail, int node_data)
{
  struct node* new_tail = create_node(node_data);
  tail->next = new_tail;
  return new_tail;
}

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