抽象数据类型的实现 (C运行)
#define ERROR 0
#define OK 1
struct STU
{ char name[20];
char stuno[10];
int age; int score;
}stu[50];
struct LIST
{ struct STU stu[50];
int length;
}L;

int printlist(struct LIST L)
{ int i;
printf("name stuno age score\n");
for(i=0;i<L.length;i++)
printf("%s %s\t%d\t%d\n", L.stu[i].name, L.stu[i].stuno, L.stu[i].age, L.stu[i].score);
printf("\n");
}

int listinsert(struct LIST *L,int i,struct STU e)
{ struct STU *p,*q;
if (i<1||i>L->length+1)
return ERROR;
q=&(L->stu[i-1]);
for(p=&L->stu[L->length-1];p>=q;--p)
*(p+1)=*p; *q=e; ++L->length;
return OK;
}/*ListInsert Before i */

main()
{ struct STU e;
L.length=0;
strcpy(e.name,"zmofun");
strcpy(e.stuno,"100001");
e.age=80;
e.score=1000;
listinsert(&L,1,e);
printlist(L);
printf("List length now is %d.\n\n",L.length);

strcpy(e.name,"bobjin");
strcpy(e.stuno,"100002");
e.age=80;
e.score=1000;
listinsert(&L,1,e);
printlist(L);
printf("List length now is %d.\n\n",L.length);
}

运行结果:

E:\ZM\Zmdoc\datastru\class02>listdemo

name stuno age score

zmofun 100001 80 1000

List length now is 1.

name stuno age score

bobjin 100002 80 1000

zmofun 100001 80 1000

List length now is 2.


请问这些代码的作用和实现的目的是什么 ?

[此贴子已经被作者于2006-3-26 22:09:38编辑过]

难道是问题太菜了 ?!
咋没有人指点指点呢 ?

哭了
还没有人帮忙啊 ?

难道人都跑水区了 ?!

写的不错啊！
恐怕好多人看不懂哦！就没有回

不是我写的啊...
我自己看不明白,才问的...

晕,哪看不懂?
结果都有了还不容易推.

#define ERROR 0
#define OK 1
struct STU
{ char name[20];
char stuno[10];
int age; int score;
}stu[50];
struct LIST
{ struct STU stu[50];
int length;
}L;

int printlist(struct LIST L) //输出函数就不用说了吧
{ int i;
printf("name stuno age score\n");
for(i=0;i<L.length;i++)
printf("%s %s\t%d\t%d\n", L.stu[i].name, L.stu[i].stuno, L.stu[i].age, L.stu[i].score);
printf("\n");
}

int listinsert(struct LIST *L,int i,struct STU e)
{ struct STU *p,*q;
if (i<1||i>L->length+1) //如果位置超出范围,抛出异常
return ERROR;
q=&(L->stu[i-1]); //将数组结构体的最后一位赋给q
for(p=&L->stu[L->length-1];p>=q;--p) 将i后结构向后复制
*(p+1)=*p; *q=e; ++L->length; 插入e至第i个位置
return OK;
}/*ListInsert Before i */

main()
{ struct STU e;
L.length=0;
strcpy(e.name,"zmofun");
strcpy(e.stuno,"100001");
e.age=80;
e.score=1000;
listinsert(&L,1,e);
printlist(L);
printf("List length now is %d.\n\n",L.length);

strcpy(e.name,"bobjin");
strcpy(e.stuno,"100002");
e.age=80;
e.score=1000;
listinsert(&L,1,e);
printlist(L);
printf("List length now is %d.\n\n",L.length);
}
就是一个简单的插入而已.