无头结点的单链表尾插法创建输不出结果

Question

#include <stdio.h>
#include <stdlib.h>
struct ListNode 
{
    int data;
    struct ListNode *next;
};
void readlist()
{
    struct ListNode *head,*p,*q;
    int i;
    head=(struct ListNode*)malloc(sizeof(struct ListNode)); 
    head->next=NULL;
    p=head;
    while(1)
    {
        scanf("%d",&i);
        if(i==-1)
        break;
        q=(struct ListNode*)malloc(sizeof(struct ListNode));
        q->next=NULL;
        q->data=i;
        q=p;
        p=p->next;
 }
 p->next=NULL;
    
 } 
 int main ()
 {
     struct ListNode *head,*p;
     readlist();
     p=head;
     while(p)
     {
         printf("%d ",p->data);
         p=p->next;
     }
 }

Answer 1

 
#include <stdio.h>
#include <stdlib.h>
struct ListNode 
{
    int data;
    struct ListNode *next;
};
struct ListNode * readlist()
{
    struct ListNode *head=NULL,*p,*q;
    int i;
    while(1)
    {
        scanf("%d",&i);
        if(i==-1)
            break;
        q=(struct ListNode*)malloc(sizeof(struct ListNode));
        q->next=NULL;
        q->data=i;
        if(head == NULL)
            p = head = q;
        else 
        {
            p->next = q;
            p = p->next;
        }
 }
 p->next=NULL;
return head;
 } 
 int main ()
 {
     struct ListNode *head,*p;
     head  = readlist();
     p=head;
     while(p)
     {
         printf("%d ",p->data);
         p=p->next;
     }
     return 0;
 }