# STRUCTURE

## STRUCTURE

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript

1. STRUCTURE 為一種複合式的資料型態，可用來把相關的資料組織 成為有意義的資料單元。 namecard： char name[10]; int age; char tel[10] char e-mail[60] 名片： name age tel e-mail

2. 結構的定義 struct namecard { char name[20]; int age; char tel[10]; char e-mail[60]; };

3. 結構的變數 struct namecard mycard; typedef struct namecard NAMECARD; NAMECARD mycard;

4. 結構的變數初始值、輸入、輸出 NAMECARD A = {“mike”, 30}; printf("Enter you name:\n"); scanf("%s", A.name); printf("Enter you age:\n"); scanf("%d", &A.age); printf("You name is %s\n", A.name); printf("You are %d years old\n", A.age);

5. 結構陣列 Mike 30 Rose 24 Lily 28 Jack 27 Tom 19 cardbook[] 0 1 2 3 4 cardbook[]={{“Mike”,30}, {“Rose”,24}, {“Lily”,28}, {“Jack”,27}, {“Tom”, 19}};

6. 結構陣列的輸入、輸出 NAMECARD cardbook[5]; for(i=0 ; i<5 ; i++) { printf("Enter you name:\n"); scanf("%s", cardbook[i].name); printf("Enter you age:\n"); scanf("%d", &cardbook[i].age); } for(i=0 ; i<5 ; i++) printf("%s,%d\n", cardbook[i].name, cardbook[i].age);

7. 結構指標 NAMECARD A={“Mike”, 30}, B={“Rose”,28}; NAMECARD *cptr; cptr = &A; printf(“nanme:%s, age:%d\n”, cptr->name, cptr->age}; cptr =&B; printf(“nanme:%s, age:%d\n”, cptr->name, cptr->age};

8. 結構指標與結構陣列 NAMECARD cardbook[3]={{"MIKE", 29}, {"COCO", 30}, {"John", 25}}; NAMECARD *cptr; cptr=cardbook; for(i=0 ; i<3 ; i++) printf("%s,%d\n", cptr[i].name, cptr[i].age);

9. 結構的動態配置 NAMECARD *cptr; clrscr(); printf("Enter the number of namecard:\n"); scanf("%d", &n); cptr=(NAMECARD*) malloc(sizeof(NAMECARD)*n);

10. Case Study : 平面座標點 struct point { int x; int y; }; . B(5,12) .C(16,8) . A(2,3)

11. Case Study : 兩點距離 float distance(POINT, POINT); . B(5,12) . A(2,3)

12. Case Study : 求重心 . B(5,12) . G . B(16,8) . A(2,3)

13. Homework : 複數 • 請定義一名為complex的複數結構，此結構裡有兩個浮點數 • r, v; • 試定義出下列四個函示 • CP complex_add(CP, CP); • CP complex_sub(CP, CP); • CP complex_mul(CP, CP); • CP complex_sub(CP, CP); • 3. 寫一主程式輸入兩個複數，計算兩複數之和、差、乘積及商

14. Case Study : Poker Game struct card { int point; int type; int status; }; typedef struct card CARD; int type[4]={5,4,3,6}; /*  ,, ,*/ char* point[13]={"2", "3", "4", "5", "6", "7", "8", "9", "10", "J","Q", "K", "A"};

15. Case Study : Poker Game 製造一副撲克牌 poker = (CARD*) malloc(sizeof(CARD)*52); for(i=0 ; i<52 ; i++) { poker[i].type = i/13; poker[i].point = i%13; }

16. Case Study : Poker Game 洗牌 CARD temp; srand(time(NULL)); for(i=0 ; i<6000 ; i++) { r = rand()%52; while((l=rand()%52)==r); temp = p[r]; p[r] = p[l]; p[l] = temp; }

17. Case Study : Poker Game printf("Enter the number of your card:"); scanf("%d", &u); printf("%c%s\n", type[poker[u].type], point[poker[u].point]); printf("I choose :\n"); scanf("%d", &i); printf("%c%s\n", type[poker[i].type], point[poker[i].point]); if(poker[u].point < poker[i].point) printf("You lose !!\n"); else if( (poker[u].point == poker[i].point) && (poker[u].type<poker[i].type) ) printf("You lose !!\n"); else printf("You Win !!\n");

18. 10 20 30 40 串列 串列的操作 . Add node . Delete node . Show list

19. 10 20 30 40 串列 .佇列 （FIFO、LIFO） .雙向鏈結DOUBLE LINK 10 20 30 40

20. 佇列 struct node { int data; struct node * next; }; typedef struct node NODE;

21. 佇列 NODE *newnode; newnode = (NODE*) malloc(sizeof(NODE)); newnode->data = data; newnode->next = NULL;

23. 10 20 30 40 10 20 30 40 佇列（LIFO） void delete(NODE**); head head *list= (*list)->next;

24. head 10 20 30 40 printf("(%x:%d)->", ptr, ptr->data); ptr = ptr->next; 佇列（LIFO） showlist(NODE*); ptr ptr

26. 10 20 30 40 10 20 30 40 佇列（FIFO） tail head tail head

27. 10 雙向鏈結 struct node{ int data; struct node* right; struct node* left; }; typedef struct node NODE;