[英]C - Doubly linked list (isEmpty)
I have doubly linked list and Ive made a function to check whether the list is empty or not. 我有双重链表,我已经创建了一个函数来检查列表是否为空。
Function code: 功能码:
int isEmpty(list *l)
{
if(l->head== NULL && l->tail== NULL)
return 1;
else
return 0;
}
List 清单
typedef struct list
{
struct element *head;
struct element *tail;
}list1;
and Im trying to use this function but there is nothing when i open console 和我试图使用此功能,但是当我打开控制台时什么也没有
case 11:
printf("List is : %d\n", isEmpty(&list1));
break;
list1
is a type so: printf("List is : %d\\n", isEmpty(&list1));
list1
是这样的类型: printf("List is : %d\\n", isEmpty(&list1));
is meaningless, 是没有意义的
Try: 尝试:
list1 l1;
.
.
.
case 11:
printf("List is : %d\n", isEmpty(&l1));
break;
It was working but this messsage was disappearing after some milliseconds. 它正在工作,但是这种混乱状态在几毫秒后消失了。 I added
system("pause");
我添加了
system("pause");
and its working now. 及其现在的工作。
You passing wrong parameter to the function. 您将错误的参数传递给函数。 you should pass paramater to isEmpty() such as isEmpty(struct list* l) or isEmpty(list1* l)
您应该将参数传递给isEmpty(),例如isEmpty(struct list * l)或isEmpty(list1 * l)
int isEmpty(struct list *l)
{
if(l->head== NULL && l->tail== NULL)
return 1;
else
return 0;
}
call the function as 将该函数称为
isEmpty(list1)
it gives correct result. 它给出正确的结果。
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