C语言栈队列实现二-十/二-八进制转换

C语言栈队列实现二-十/二-八进制转换

2015-04-05 Lover雪儿

 

 1 //利用栈来求取二进制数的十进制与八进制的结果

 2 #include <stdio.h>

 3 #include <stdlib.h>

 4 #include <math.h>

 5 

 6 #define STACK_INIT_SIZE        20    //初始栈大小

 7 #define STACK_INCREMENT        10  //扩充栈时每次增加的内存

 8 

 9 typedef char ElemType;            //栈中的数据类型

10 typedef struct {

11     ElemType *base;

12     ElemType *top;

13     int stackSize;

14 }sqStack;

15 

16 //初始化栈

17 void init_stack(sqStack *s){

18     s->base = (ElemType *)malloc(STACK_INIT_SIZE * sizeof(ElemType));

19     if(!s->base)

20         exit(-1);

21     s->top = s->base;

22     s->stackSize = STACK_INIT_SIZE;

23 }

24 //入栈

25 void push(sqStack *s,ElemType e){

26     if(s->top - s->base >= s->stackSize){    //检测栈是否已满

27         //若是已满，则自动扩大栈空间

28         s->base = (ElemType *)realloc(s->base, (s->stackSize + STACK_INCREMENT)*sizeof(ElemType));

29         if(!s->base)

30             exit(-1);

31     }

32     *(s->top) = e;

33     s->top++;

34 }

35 //出栈

36 void pop(sqStack *s, ElemType *e){

37     if(s->top == s->base){

38         return ;

39     }

40     *e = *(--(s->top));

41 }

42 //求栈数据的个数,由于我们不会对其修改，故此处不需传指针

43 int  stack_len(sqStack s){

44     return (s.top - s.base); //返回数据的个数    

45 }

46 

47 int main(void){

48     ElemType c;

49     sqStack s;

50     int len = 0, i = 0, sum_10 = 0;

51     int j = 0, sum_8 = 0,tmp_8 = 0;

52     

53     init_stack(&s);        //初始化栈

54     while(1){

55         printf("请输入二进制数，输入#符号表示结束，开头输入q/Q退出!\n");

56         scanf("%c",&c);    //按字符格式接收

57         if(c == 'q' || c == 'Q')

58             break;

59         while(c != '#'){

60             push(&s,c);        //入栈

61             scanf("%c",&c);

62         }

63         getchar();        //将回车从当前缓冲区去除，用户按下回车后结束

64         

65         len = stack_len(s);        //获取栈的当前容量

66         printf("栈的当前容量是: %d\n",len);

67 

68         for(i = 0; i<len ; i++){

69             pop(&s, &c);  //传值使用引用方式

70             sum_10 =(int)( sum_10 + (c-48) * pow(2 , i));    //将ASCII转换为整形

71 

72             tmp_8 =(int)( tmp_8 + (c-48) * pow(2 , i%3));    //将ASCII转换为整形

73 

74             //printf("%d  %d\n",i%4,tmp_8);

75             if(i%3 == 2){        

76                 sum_8 += (int)(tmp_8 * pow(10 , (j++)));

77                 tmp_8 = 0;

78             }    

79         }

80         if(tmp_8 != 0){

81             sum_8 += (int)(tmp_8 * pow(10 , (j++)));

82             tmp_8 = 0;

83             j = 0;

84         }

85         printf("转换为十进制数是: %d 八进制数: 0%d\n",sum_10,sum_8);

86         sum_10 = 0;

87         sum_8 = 0;

88         j = 0;

89     }

90     return 0;

91 }