c - LinkedList 堆栈损坏

Question

#define _CRT_SECURE_NO_WARNINGS

#include <stdio.h>
#include <stdlib.h>
typedef struct listnode{
    int item;
    struct listnode *next;
}ListNode;

typedef struct _linkedlist{
    ListNode *head;
    int size;
} LinkedList;

void printList(LinkedList *ll);
int sizeList(LinkedList *ll);
int insertSorted(LinkedList *ll, int value);
int removeDuplicates(LinkedList *ll);

int main()
{
    int choice, i = 0;
    ListNode *temp=NULL;
    LinkedList *ll=NULL;

    printf("1. create LinkedList\n2. insertSorted\n3. removeDuplicates\nChoose an option: ");
    scanf("%d", &choice);
    switch (choice)
    {
    case 1:
        printf("Enter a list of numbers, terminated by the value -1: ");
        scanf(" %d", &i);
        while (i != -1){
            if (ll == NULL)
            {
                ll = malloc(sizeof(LinkedList));
                temp = ll;
            }
            else
            {
                temp->next = malloc(sizeof(ListNode));
                temp = temp->next;

            }
            temp->item = i;
            scanf(" %d", &i);
        }
        temp->next = NULL;
        printList(&ll);
        printf("Size of linked is %d", sizeList(&ll));
        break;
    case 2:

    default:
        break;
    }
}

void printList(LinkedList *ll)
{
    ListNode *temp = ll->head;
    if (temp == NULL)
        return;
    while (temp!=NULL)
    {
        printf("%d ", temp->item);
        temp = temp->next;
    }
    printf("\n");
}

int sizeList(LinkedList *ll)
{
    int size=0;

    ListNode *temp = ll->head;
    if (temp == NULL)
        return 0;

    while (temp != NULL)
    {
        size++;
        ll->size = size;
        temp = temp->next;
    }
    return ll->size;
}

我想创建一个链表并计算链表的大小并输出。我设法获取大小并打印出列表，但最后，我的程序显示调试错误并指出运行时检查失败#2 - 变量“ll”周围的堆栈已损坏。我可以知道为什么会发生这种情况吗？

Answer 1

其中一个不正确的地方是您的 main() 函数。

if (ll == NULL)
{
    ll = malloc(sizeof(LinkedList));
    temp = ll; // <- This is incorrect!!
}
else
{
    temp->next = malloc(sizeof(ListNode));
    temp = temp->next;
}

temp 是一个ListNode，如何为其分配一个LinkedList？同样，对于这一行，temp 和 ll 都指向同一内存，并且在其中一个上进行操作将覆盖其他。

它可能应该是这样的:

if (ll == NULL)
{
    ll = malloc(sizeof(LinkedList));
    temp = malloc(sizeof(ListNode));
    temp->next = NULL;
    ll->head = temp;
    ll->size = 1;
}
else
{
    temp->next = malloc(sizeof(ListNode));
    temp = temp->next;
}

但这也没有想象中那么美好。我更愿意:

int main()
{
    int choice, i = 0;
    ListNode *temp=NULL;
    LinkedList ll = { NULL, 0 };

    printf("1. create LinkedList\n2. insertSorted\n3. removeDuplicates\nChoose an option: ");
    scanf("%d", &choice);
    switch (choice)
    {
    case 1:
        printf("Enter a list of numbers, terminated by the value -1: ");
        scanf(" %d", &i);
        while (i != -1){
            if (ll.head == NULL)
            {
                temp = malloc(sizeof(ListNode));
                ll.head = temp;
            }
            else
            {
                temp->next = malloc(sizeof(ListNode));
                temp = temp->next;
            }
            temp->item = i;
            scanf(" %d", &i);
        }
        temp->next = NULL; // <- Please take a second look at this line. What happens if your first entry is -1?
        printList(&ll); // <- This too...
        printf("Size of linked is %d", sizeList(&ll)); // <- and this as well...
        break;
    case 2:

    default:
        break;
    }
}

这是因为不需要动态分配LinkedList。只要程序正在运行，它就一直存在，因此将其放在堆上是没有意义的。

您的代码还存在其他问题，我建议您使用橡皮鸭代码，并强烈建议阅读 this link on debugging