#include <stdio.h>
#define NULL 0
struct xuesheng
{
    int num;
    float score[3];
    struct xuesheng *next;
};
  struct xuesheng * heada(void)
{
    struct xuesheng l,m,n,*headx,*xp;
    l.num=1;l.score[0]=90;l.score[1]=90;l.score[2]=90;
    m.num=3;m.score[0]=80;m.score[1]=80;m.score[2]=80;
    n.num=5;n.score[0]=70;n.score[1]=70;n.score[2]=70;
    headx=&l;
    l.next=&m;
    m.next=&n;
    n.next=NULL;
    putchar('\n');
    printf("输出函数1的链表\n");
    for(xp=headx;xp!=0;xp=xp->next)
    printf("%d,%f,%f,%f\n",xp->num,xp->score[0],xp->score[1],xp->score[2]);
    putchar('\n');
    putchar('\n');
    return(headx);
}
  struct xuesheng * headb(void)
{
    struct xuesheng o,p,q,*heady,*xp;
    o.num=2;o.score[0]=90;o.score[1]=90;o.score[2]=90;
    p.num=4;p.score[0]=80;p.score[1]=80;p.score[2]=80;
    q.num=6;q.score[0]=70;q.score[1]=70;q.score[2]=70;
    heady=&o;
    o.next=&p;
    p.next=&q;
    q.next=NULL;
    printf("输出链表2\n");
    for(xp=heady;xp!=NULL;xp=xp->next)
    printf("%d,%f,%f,%f\n",xp->num,xp->score[0],xp->score[1],xp->score[2]);
    putchar('\n');
    putchar('\n');
    return(heady);
}
  main()
  {   int i,j,n,v;
      float f0;
      struct xuesheng  zp[6];
      struct xuesheng *headnew,*p1,*p2;
       p1=heada();
      for(i=3;i<6;i++)
      {
         zp[i].num=p1->num;
         for(v=0;v<3;v++)
         zp[i].score[v]=p1->score[v];
          p1=p1->next;
      }
      p1=headb();
      for(i=3;i<6;i++)
      {
          zp[i].num=p1->num;
         for(v=0;v<3;v++)
         zp[i].score[v]=p1->score[v];
         p1=p1->next;
      }
      
      for(j=0;j<5;j++)
          for(i=0;i<5-j;i++)
          if((zp[i]).num>(zp[i+1]).num)
          {   
              n=zp[i].num;
              zp[i].num=zp[i+1].num;
              zp[i+1].num=n;
              for(v=0;v<3;v++)
              { f0=zp[i].score[v];
              zp[i].score[v]=zp[i+1].score[v];
              zp[i+1].score[v]=f0;}
          }
        
      headnew=zp;
      for(i=1;i<=4;i++)
      zp[i].next=&zp[i+1];
      zp[5].next=NULL;
      for(p2=headnew;p2!=NULL;p2=p2->next)
          printf("%d,%f,%f,%f\n",p2->num,p2->score[0],p2->score[1],p2->score[2]);
给的条件是2个有3个结点的链表。算法为：先将2个链表放置在以个结构体数组中，再对数组元素进行排序，编译时没有错误，运行时出现
以下提示：好像是内存不能被重置什么的，请有经验的同志给点提示。

那么如何将函数中的地址送达主函数呢？算法应该没问题吧。