C语言动态链表问题--链表头

Question

第一个程序中head指向a是链表头

#include<stdio.h>
#include<malloc.h>
#include<stdlib.h>
struct student
{
    int data;
    struct student *next;
}a;
struct student *p,*q,*head;
void f(struct student *L,int n)
{
    int t,i,j;
    p=L;
    q = p->next;
    for(i=0;i<n;i++)
    {
        for(j=0;j<n;j++)
        {
            if(p->data < q->data) 
            {
                t = p->data;
                p->data = q->data;
                q->data = t;
            }
            p = q;
            q = q->next;
        }
        p = L;
        q = p->next;
    }
}
int main()
{
    struct student *p1;
    p1=&a;
    int i,n,c=0;
    printf("请输入n:\n");
    scanf("%d",&n);
    head=q=&a;
    printf("\n");
    q->next=NULL;
    printf("请输入%d个同学的成绩：\n",n);
    for(i=0;i<n;i++)
    {
        p = (struct student*)malloc(sizeof(struct student));
        scanf("%d",&p->data);
        q->next = p;
        q = q->next;
    }
    f(p1,n);
    printf("\n");
    for(i=0;i<n;i++)
    {
        printf("%d\n", head->data);
        head = head->next;
    }
    return 0;
}
 

第二个程序中head同样指向a，但是会丢掉一个数据，为什么

#include<stdio.h>
#include<malloc.h>
#include<stdlib.h>
struct student
{
    int data;
    struct student *next;
}a;
struct student *p,*q,*head;
int main()
{
    struct student *p1;
    p1=&a;
    head=&a;
    int i,n,c=0;
    printf("请输入n:\n");
    scanf("%d",&n);
    q=&a;
    printf("\n");
    q->next=NULL;
    printf("请输入%d个同学的成绩：\n",n);
    for(i=0;i<n;i++)
    {
        p = (struct student*)malloc(sizeof(struct student));
        scanf("%d",&p->data);
        if(p->data == -1)
        {
            printf("输入结束\n");
            p -> next = NULL;
            break;
        }
        q->next = p;
        q = q->next;
    }
    p->next=NULL;
    printf("\n");
    printf("打印数据：\n");
    for(i=0;head!=NULL;i++)
    {
        printf("%d\n", head->data);
        head = head->next;
    }
    return 0;
}
 

Answer 1

因为打印的个数不一样

你把上面改成n+1就可以了