LinkedList 长度、访问和通过堆实现。 C

Question

我正在尝试从 main.c 文件打印我在另一个名为 linklist.c 的 .c 文件中创建的链表的长度。它不起作用，我相信这与指针和/或内存管理有关。我主要在这里质疑堆。一些指导将不胜感激。

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

int main()
{
    printf("Hello world!\n");

    struct node* mylist = BuildOneTwoThree();
    int length = Length(mylist);

    printf(mylist->data);
    printf(length);
    return 0;
}

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


// Return the number of nodes in a list (while-loop version)
int Length(struct node** head) {
   int count = 0;
   struct node* current = head;
   while (current != NULL) {
      count++;
      current = current->next;
   }
   return(count);
}

/*
 Build the list {1, 2, 3} in the heap and store
 its head pointer in a local stack variable.
 Returns the head pointer to the caller.
*/
struct node* BuildOneTwoThree() {
   struct node* head = NULL;
   struct node* second = NULL;
   struct node* third = NULL;
   head = malloc(sizeof(struct node)); // allocate 3 nodes in the heap
   second = malloc(sizeof(struct node));
   third = malloc(sizeof(struct node));
   head->data = 1; // setup first node
   head->next = second; // note: pointer assignment rule
   second->data = 2; // setup second node
   second->next = third;
   third->data = 3; // setup third link
   third->next = NULL;
   // At this point, the linked list referenced by "head"
   // matches the list in the drawing.
   return head;
}

/*
 Takes a list and a data value.
 Creates a new link with the given data and pushes
 it onto the front of the list.
 The list is not passed in by its head pointer.
 Instead the list is passed in as a "reference" pointer
 to the head pointer -- this allows us
 to modify the caller's memory.
*/
void Push(struct node** headRef, int data) {
   struct node* newNode = malloc(sizeof(struct node));
   newNode->data = data;
   newNode->next = *headRef; // The '*' to dereferences back to the real head
   *headRef = newNode; // ditto head points to new node
}

// Given a list and an index, return the data
// in the nth node of the list. The nodes are numbered from 0.
// Assert fails if the index is invalid (outside 0..lengh-1).
int GetNth(struct node* head, int index) {
    struct node* current = head;
    int answer = 0;
    int x = index;
    if(x <= 0 || x >= sizeof(head)-1 )
        {
            return -1;
        }
    for(int i = 0; i <= sizeof(head)-1; i++){
        if (i == x){
            return current->data;
        }
        current = current->next;
    }
}

如您所见，我使用 BuildOneTwoThree 函数来构建链接列表并正在编写适当的函数...当我尝试将 mylist 访问到输出时它崩溃了。

Answer 1

在大多数情况下，代码似乎仅从问题的角度起作用，printf 必须正确格式化。

printf("%i", length);

Answer 2

1. 需要类型转换 malloc
2. 在函数参数长度中将 **head 更改为 *head
3. 使用正确的 printf 语句。

这个问题仅限于打印链接列表的长度。请在下面找到代码：

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

struct node
{
    int data;
    struct node *next;
};

struct node* BuildOneTwoThree();
int Length(struct node* head);
int main()
{
    printf("Hello world!\n");

    struct node* mylist = BuildOneTwoThree();
    int length = Length(mylist);

    printf("data =%d\n",mylist->data);
    printf("length = %d",length);
    return 0;
}


// Return the number of nodes in a list (while-loop version)
int Length(struct node* head) {
   int count = 0;
   struct node* current = head;
   while (current != NULL) {
      count++;
      current = current->next;
   }
   return(count);
}

/*
 Build the list {1, 2, 3} in the heap and store
 its head pointer in a local stack variable.
 Returns the head pointer to the caller.
*/
struct node* BuildOneTwoThree() {
   struct node* head = NULL;
   struct node* second = NULL;
   struct node* third = NULL;

   head = (struct node *)malloc(sizeof(struct node)); // allocate 3 nodes in the heap
   second = (struct node *)malloc(sizeof(struct node));
   third = (struct node *)malloc(sizeof(struct node));

   head->data = 1; // setup first node
   head->next = second; // note: pointer assignment rule
   second->data = 2; // setup second node
   second->next = third;
   third->data = 3; // setup third link
   third->next = NULL;
   // At this point, the linked list referenced by "head"
   // matches the list in the drawing.
   return head;
}