C语言 打印链表时出现分段错误

Question

我一直在阅读斯坦福大学关于链表的教程。我使用了其中一个函数来创建三个数字（1,2,3）的列表。函数本身不会打印结果，所以我决定自己测试它。但是，当我运行它时，它给了我分段错误。

话虽如此，当我删除函数并将代码复制到 main 时，它可以工作。有人可以解释为什么主要功能不起作用吗？

这是给我分段错误的代码：

#include <stdio.h>
#include <stdlib.h>
      struct node {
         int            data;
         struct node*   next;
};
struct node* BuildOneTwoThree() 
{
   struct node* head = NULL;
   struct node* second = NULL;
   struct node* third = NULL;
   head = malloc(sizeof(struct node));
   second = malloc(sizeof(struct node));
   third = malloc(sizeof(struct node));
head->data = 1;
head->next = second;
second->data = 2;
second->next = third;
third->data = 3;
third->next = NULL;
return head;
}

int main()
{
   struct node* head;
   struct node* second;
   struct node* third;
   struct node*   next;
   int data;
   BuildOneTwoThree();
   struct node* current = head;

while(current != NULL)
   {
      printf("%d ", current->data );
      current= current->next;
   }
}

这个工作：

#include <stdio.h>
#include <stdlib.h>
      struct node {
         int            data;
         struct node*   next;
};

int main()
{
   int data;
   struct node* head = NULL;
   struct node* second = NULL;
   struct node* third = NULL;
   head = malloc(sizeof(struct node));
   second = malloc(sizeof(struct node));
   third = malloc(sizeof(struct node));
head->data = 1;
head->next = second;
second->data = 2;
second->next = third;
third->data = 3;
third->next = NULL;
   struct node* current = head;

while(current != NULL)
   {
      printf("%d ", current->data );
      current= current->next;
   }
}

Answer 1

在不工作的版本中，您将忽略 BuildOneTwoThree 的返回值，并将未初始化的局部变量从 main head（与BuildOneTwoThree作用域中同名的局部变量不同）分配给变量current。

因此，打印代码应使用：

struct node* head = BuildOneTwoThree();
current = head;

相反，利用在 BuildOneTwoThree() 中分配的head节点，并分配给main的头指针。