void delate(struct student *head,int n)
  {
      struct student *p,*q,*p1;
      printf("plase input delate student old:");
      scanf("%d",&n);
      p=head;
      while(p->next!=NULL)
      {
           if(head->old==n)
          {
              p1=head;
             head=head->next;
            p=head;
        //    head->next=p->next;
            free(p1);
              break;
          }
          else if(p->next->old==n)
          {
              q=p->next;
              p->next=q->next;
             free(q);
             break;
          }
         
         
       else
           p=p->next;
      
  }
      
}
为什么编译后返回就错了呢，是因为没free吗

[此贴子已经被作者于2016-6-5 16:53编辑过]

因为在delate?函数里的head实际上是一个按值传递的指针，你在函数里改head不能影响外面的"head"
而且这个逻辑也有点问题，如果只有链表中只有一个人的时候那就什么也不会做了

[quote]以下是引用浅暗花璃在2016-6-5 16:51:43的发言：

void delate(struct student *head,int n)//void 改成struct student
  {
      struct student *p,*q,*p1;
      printf("plase input delate student old:");
      scanf("%d",&n);//n是实参传递过来的，是有值，你这里再重新输入是不是有问题。
    /*if(head==NULL)
        {
            printf("空链表\n");
            return head;
            }最好对链表进行判断是否空。*/
      p=q=head;
      while(p->next!=NULL)
      {
        if(head->old==n)
          {
          p1=head;
             head=head->next;
            p=head;
        //    head->next=p->next;
            if(p->old==n)
            {
                if(p==head) head=p->next;
                else q->next=p->next;
              free(p);
              break;
          }
          else if(p->next->old==n)
          {
              q=p->next;
              p->next=q->next;
             free(q);
             break;
          }
         
         
       else
           p=p->next;
      
    q=p;
    p=p->next;
  }
     return head;
}

[此贴子已经被作者于2016-6-5 20:00编辑过]

void delate(struct student *head,int n)
   {
       struct student *p,*q,*p1;
       printf("plase input delate student old:");
       scanf("%d",&n);//n是输入的话，就不用放在函数形参里了，在函数体里定义就可以了
       p=head;
       while(p->next!=NULL)
       {
            if(head->old==n)
           {
               p1=head;
              head=head->next;
             p=head;
         //    head->next=p->next;
             free(p1);
               break;
           }
           else if(p->next->old==n)
           {
               q=p->next;
               p->next=q->next;
              free(q);
              break;
           }
           
         
        else
            p=p->next;
        
   }
      
 }