依自己理解写了双链表，运行好像行，请大家指点一下,该如何处理

依自己理解写了双链表，运行好像行，请大家指点一下
#include<iostream>
#include<string>
using namespace std;
struct student
{
char name[20];
float score;
student *prior;
student *next;
};
void main()
{
student *head=NULL,*tail=NULL,*pnew,*p=NULL;
char name[20];
float score;
while(1)
{
    cout<<"input name and score:(0 to exit)"<<endl;
cin>>name;
if(name[0]=='0')break;
cin>>score;

pnew=new student;
strcpy(pnew->name ,name);
pnew->score=score;
pnew->prior=p;
pnew->next=NULL;

if(tail==NULL)
          head=pnew;
else
tail->next=pnew;
        tail=pnew;
p=pnew;

}
student *pos=tail;
while(pos!=NULL)
{
     cout<<pos->name<<" "<<pos->score;
 pos=pos->prior ;
 if(pos!=NULL)cout<<"-->";
}
    cout<<endl; 
while(head!=NULL)
{
pos=head;
head=head->next ;
delete pos;
}


}
------解决方案--------------------

#include<iostream>                                                                                                                                                               

#include<string.h>

#include <stdio.h>

using namespace std;

typedef struct student

{

    char name[20];

    float score;

    struct student *prior;

    struct student *next;

}Student;


int main()

{

    Student *head=NULL,*tail=NULL,*pnew,*p=NULL;

    char name[20];

    float score;

    while(1)

    {   

            cout<<"input name and score:(0 to exit)"<<endl;

            cin>>name;

            if( name[0] =='0')

                break;

            cin>>score;


            pnew=new student;

            strcpy(pnew->name ,name);

            pnew->score=score;

            pnew->prior=p;

            pnew->next=NULL;


            if(tail==NULL)

                head=pnew;

            else

                tail->next=pnew;

            tail=pnew;

            p=pnew;

    }   

 Student *pos=tail;

    while(pos!=NULL)

    {

        cout<<pos->name<<" "<<pos->score;

        pos=pos->prior ;

        if(pos!=NULL)

            cout<<"-->";

    }

    

    cout<<endl; 

    while(head!=NULL)

    {

        pos=head;