题目：有2个链表a，b，设结点中包含学号、姓名。从a链表中
删去和b链表中相同的学号的那些结点。

自己写的代码：
#include "stdio.h"
#include "malloc.h"

void print(struct student *head);
struct student *creat();
struct student *del(struct student *head,long num);


struct student
{
    int num;
    char name;
    struct student *next;
};
int n;

void main()
{
    struct student *p1,*p2,*p3;
    p1=creat();
    p2=creat();
    for(p2;p2->next!=NULL;p2++)
    p3=del(p1,p2->next->num);
    print(p3);
}


struct student *creat()
{
    struct student *head,*p1,*p2;
    int n;
    p1=p2=(struct student*)malloc(sizeof(struct student));
    cin>>p1->num>>p1->name;
    head=NULL;
    while(p1->num!=0)
    {
        n=n+1;
        if(n==1)head=p1;
        else p2->next=p1;
        p2=p1;
        p1=(struct student *)malloc(sizeof(struct student));
        cin>>p1->num>>p1->name;

    }
    p2->next=NULL;
    return head;
}

void print(struct student *head)
{
    struct student *p;
    p=head;
    cout<<"Now,These "<<n<<"records are:";
    if(head!=NULL)
    {

        do
        {
            cout<<p->num<<p->name;
            p=p->next;
        }while(p!=NULL);
    }
}

struct student *del(struct student *head,long num)
{
    struct student *p1,*p2;
    if(head==NULL){cout<<"List null";}
    p1=head;
    while(num!=p1->num&&p1->next!=NULL)
    {
        p2=p1;
        p1=p1->next;
    }
    if(num==p1->num)
    {
        if(p1==head)
            head=p1->next;
        else p2->next=p1->next;
        cout<<"delete "<<num;
        n=n-1;
    }
    else cout<<num<<"not been found!";
    return head;

}
运行和编译都通过了，但是自己逐步调试的时候发现运行到for(p2;p2->next!=NULL;p2++)；的时候出现错误。自己主要是想通过这一句p2=creat();将num调出来，以便删除函数p3=del(p1,p2->next->num);能够删除p1=creat();中和p2=creat();相同num的结点，而达到题目的要求。