编写一个输出链表的函数print。

解：程序：

#include<stdio.h>

#include<stdlib.h>

#define LEN sizeof(struct Student)

struct Student

{

long num;

float score;

struct Student *next;

};

int n;

struct Student *creat()//建立链表的函数

{

struct Student *head;

struct Student *p1, *p2;

n = 0;

p1 = p2 = (struct Student *)malloc(LEN);

scanf("%ld,%f", &p1->num, &p1->score);

head = NULL;

while (p1->num != 0)

{

n = n + 1;

if (n == 1)

{

head = p1;

}

else

{

p2->next = p1;

}

p2 = p1;

p1 = (struct Student *)malloc(LEN);

scanf("%ld,%f", &p1->num, &p1->score);

}

p2->next = NULL;

return(head);

}

void print(struct Student *head)//输出链表的函数

{

struct Student *p;

printf("\nNow,These %d records are:\n",n);//records记录

p = head;//p指向第1个结点

if (head != NULL)//若不是空表

{

do

{

printf("%ld   %5.1f\n", p->num, p->score);

p = p->next;

} while (p != NULL);

}

}

void main()

{

struct Student *head;

head= creat();//调用函数返回链表第1个结点的地址

print(head);

}

结果：

1001,77

1003,88

1005,99

0,0

 

Now,These 3 records are:

1001    77.0

1003    88.0

1005    99.0

请按任意键继续. . .

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