#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。我设法获得大小并打印出列表,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这样吗?
答案 0 :(得分:1)
其中一个不正确的事情是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ListNode
,如何为其分配LinkedList
?同样,对于该行,temp
和ll
都指向相同的内存,并且在一行上操作覆盖另一行。
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;
}
但这也不是那么好。我更喜欢:
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;
}
}
这是因为无需动态分配LinkedList
。只要程序正在运行,它就会存活,因此没有必要将它放在堆上。
您的代码还存在其他问题,我建议您使用 rubber ducking 代码强烈建议您阅读this link on debugging