程序代码：

# include <stdio.h>
# include <malloc.h>
# include <stdlib.h> 


//函数前置声明
struct Student * InputStudent(int);
void OutputStudent(struct Student *);
void paixu(struct Student *);


//学生结构体
struct Student
{
    int a;        //年龄
    char b[10];   //姓名
    struct Student * pNext; //指针域
};


//主函数
int main(void)
{
    int len;
    printf("请输入学生的人数：");
    scanf("%d", &len);
   
    struct Student * p;
    p = InputStudent(len);
    printf("\n\n");
    paixu(p);
   
    OutputStudent(p);
    return 0;
}


//输入函数
struct Student * InputStudent(int len)
{
    int i;
   
    //构建头结点
    struct Student * p = (struct Student *)malloc(sizeof(struct Student));
   
    if( NULL == p)
    {
        printf("构建失败！");
        exit(-1);
    }
       
   
     //构建链表
     struct Student * t = p;
     t->pNext = NULL;
    
    for(i=0; i<len; i++)
    {
        struct Student * pNEW = (struct Student *)malloc(sizeof(struct Student));
        if( NULL == pNEW)
        {
            printf("构建失败！");
            exit(-1);
        }
       
        printf("请输入第 %d 个学生的信息：\n", i+1);
        printf("年龄：");
        scanf("%d", &pNEW->a);
        printf("姓名：");
        scanf("%s", pNEW->b);
       
        t->pNext = pNEW;
        pNEW->pNext = NULL;
        t = pNEW;
       
    }
   
    return p;
}

//输出函数
void OutputStudent(struct Student * p)
{
    int i = 1;
    p = p->pNext;
   
    while( NULL != p)
    {
        printf("第 %d 个学生的信息：\n", i);
        printf("年龄：%d", p->a);
        printf("姓名：%s\n", p->b);
        p = p->pNext;
        i++;
    }
    return;
}


//排序函数
void paixu(struct Student * p)
{
    struct Student * t;
    struct Student * t1;
    struct Student * t2;
    struct Student tt;
    int z = 0;
    int i, j;
    p = p->pNext;//取首节点
    t = t2 = p;
   
    //取链表数
    while( NULL != t)
    {
        z++;
        t = t->pNext;
    }
   
    t = p->pNext;//取第二节点
   
    //开始循环比较
    for(i=0; i<z; i++)
    {
        for(j=0; j<z-i; j++)
        {
            if( (p->a) < (t->a) )
            {
                tt = *p;
                *p = *t;
                *t = tt;
   
                t1 = p->pNext;
                p->pNext = t->pNext;
                t->pNext = t1;
            }
           
           
           
            //进入下一个节点
            p = p->pNext;
            t = t->pNext;

            //获取首节点地址进入第二次的i循环
            if( NULL == t) 
            {
                p = t2;
                t = p->pNext;
            }
        }
    }
    return;
}