用指針處理鏈表

//用指針處理鏈表
//建立一個簡單的鏈表，它由3個學生數據的結點組成，要求輸出各結點中的數據
/*#include <stdio.h>
struct Student
{
	int num;
	float score;
	struct Student *next;
};
int main()
{
	struct Student a,b,c,*head,*p;
	a.num = 10101;
	a.score = 89.5;
	b.num = 10103;
	b.score = 90;
	c.num = 10107;
	c.score = 85;
	head = &a;							//將結點a的起始地址賦給頭指針head
	a.next = &b;							//將結點b的起始地址賦給a結點的next成員
	b.next = &c;							//將結點c的起始地址賦給b結點的next成員
	c.next = NULL;							//將結點c的next成員不存放其他結點地址
	p = head;							//使p也指向a結點
	do
	{
		printf("%1d%5.1f\n",p->num,p->score);
		p = p->next;						//使p指向下一結點
	}while(p != NULL);						//輸出完c結點後p的值爲NULL，循環終止
	return 0;
}*/
//寫一函數建立一個有3名學生數據的單向動態鏈表
#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(void)
{
	struct Student * head;
	struct Student *p1,*p2;
	n = 0;
	p1 = p2 = (struct Student *)malloc(LEN);			//開闢一個新單元
	scanf("%ld,%f",&p1->num,&p1->score);				//輸入第1個學生的學號和成績
	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);
	}
	p1->next = NULL;
	return (head);
}
int main()
{
	struct Student *pt;
	pt = creat();
	printf("\nnum:%ld\nscore:%5.1f\n",pt->num,pt->score);
	return 0;
}