为什么当 current->next == NULL 时 current = current->next 会出现分段错误?

Why current = current->next gives segmentation fault when current->next == NULL?

我有下面的 C 程序来实现升序链表。问题出在 buildList() 函数中,因此您可以忽略除 main() 和 buildList() 之外的其他函数。

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

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

void buildList(struct node **, int);
void printList(struct node **);

int main()
{
    struct node *head;
    head = NULL; /*Empty list*/

    int hold, i, j;
    printf("How many integers in this list: ");
    scanf("%d",&i);
    j = 1;
    for(i; i > 0; i--)
    {
            printf("Integer %d: ",j);
            scanf("%d",&hold);
            buildList(&head, hold);
            j++;
    }
    printList(&head);
    return 0;
}

void buildList(struct node **headRef, int data)
{
    struct node *newNode;
    newNode = malloc(sizeof(struct node));
    struct node *current, *current1;
    current = *headRef;
    //If list is empty, add the number as the first node.
    if (current == NULL)
    {
            newNode->data = data;
            newNode->next = *headRef;
            *headRef = newNode;
    }
   //If the list is not empty.
    else
    {
            //If the number is not greater than first number.
            if (data <= current->data)
            {
                    newNode->data = data;
                    newNode->next = current;
                    *headRef = newNode;
            }
            //If the number is greater than the first number in list.
            else
            {
                    int flag = 0;
                    while (data > (current->data))
                    {
                            current1 = current;
                            current = current->next;
                    }


                    newNode->data = data;
                    current1->next = newNode;
                    newNode->next = current;
            }
    }
}

//Prints nodes and total number of nodes.
void printList(struct node **headRef)
{
    int count = 0;
    struct node *current;
    current = *headRef;
    while (current != NULL)
    {
            count++;
            printf("%d ",current->data);
            current = current->next;

    }
    printf("\nTotal nodes: %d\n",count);
}

在我给出一个比列表中存在的任何数字都大的数字之前,该程序工作正常。在那种情况下,我会遇到分段错误。

案例一(精细正确输出)

-bash-4.1$ ./a.out
How many integers in this list: 3
Integer 1: 5
Integer 2: 1
Integer 3: 4
1 4 5
Total nodes: 3
-bash-4.1$

情况二(此处代码中断(段错误))

-bash-4.1$ ./a.out
How many integers in this list: 3
Integer 1: 5
Integer 2: 6
Segmentation fault
-bash-4.1$

经过很长时间试图弄清楚我的代码出了什么问题,我发现当一个数字大于列表中的任何数字时,在这种情况下,它应该被插入到列表,此语句(在函数 buildList() 的最后一个 else 中)导致了问题:

  current = current->next;

我终于弄明白当current->next 为NULL 时,这条语句会导致seg 错误。

我想出了一个如下所示的解决方法,它可以提供正确的输出。

 else
 {
     int flag = 0;
     while (data > (current->data))
     {
         current1 = current;
         if(current->next != NULL)
         {
             current = current->next;
         }
         else
         {
             flag = 1;
             break;
         }
    }
    newNode->data = data;
    current1->next = newNode;
    if (flag == 1)
    {
       newNode->next = NULL;
    }
    else
    {
       newNode->next = current;
    }
}

现在我得到了正确的输出。

-bash-4.1$ ./a.out
How many integers in this list: 3
Integer 1: 5
Integer 2: 1
Integer 3: 10
1 5 10
Total nodes: 3
-bash-4.1$

现在我想知道为什么不是 current = current->next;当 current->next 为 NULL 时工作。我期待这条语句将 NULL 分配给 current.

谁能告诉我这是什么原因? 我的解决方法也很好吗?或者有更好的方法吗?

抱歉问了这么长的问题,但花了大约 2 个小时调试这个,我想我快疯了。

谢谢。

您的调试技术有点误会了您。问题出在这个循环中:

while (data > (current->data))
{
    current1 = current;
    current = current->next;
}

但不是你认为的那样。如果 current->nextNULL,则 current 会像您描述的那样设置为 NULL。然而,下一个发生的操作是循环条件——取消引用 current。布拉莫。你可能想要这样的东西:

while (current && (data > current->data))
{
    current1 = current;
    current = current->next;
}

我没有分析你的整个程序,所以除了明显的崩溃之外,我无法真正评论这部分的正确性。

当第二个输入大于第一个输入时,这个while循环不会停止。

     while ( data > (current->data))
     {
        current1 = current;
        current = current->next;
     }

您最终取消了对 NULL 指针的引用。您需要:

     while ( current != NULL && data > (current->data))
     {
        current1 = current;
        current = current->next;
     }