数据结构趣题——二进制/八进制转换

#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#define STACK_INIT_SIZE 20
#define STACKINCREMENT 10

typedef  char ElemType;
typedef struct {
    ElemType *base;
    ElemType *top;
    int stacksize;
} sqStack;

void initStack(sqStack *s)
{
    /*内存中开辟一段连续空间作为栈空间，首地址赋值给s->base*/
    s->base = (ElemType *)malloc(STACK_INIT_SIZE * sizeof(ElemType));

    if(!s->base) exit(0);     /*分配空间失败*/

    s->top = s->base;       /*最开始，栈顶就是栈底*/
    s->stacksize = STACK_INIT_SIZE;   /*最大容量为STACK_INIT_SIZE */
}

void Push(sqStack *s, ElemType e) {
    if(s->top - s->base >= s->stacksize) {
        /*栈满，追加空间*/
        s->base = (ElemType *)realloc(s->base, (s->stacksize +
                                                STACKINCREMENT) * sizeof(ElemType));

        if(!s->base) exit(0);   /*存储分配失败*/

        s->top = s->base + s->stacksize;
        s->stacksize = s->stacksize + STACKINCREMENT; /*设置栈的最大容量*/
    }

    *(s->top) = e;  /*放入数据*/
    s->top++;
}

void Pop(sqStack *s , ElemType *e) {
    if(s->top == s->base) return;

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

int StackLen(sqStack s) {
    return (s.top - s.base) ;
}

int main()
{
    ElemType c;
    sqStack s1;
    sqStack s2;
    int len, i, j, sum = 0;
    initStack(&s1);  /*创建一个栈s1，用来存放二进制字符串*/

    printf("Please input a binary number and type '#' for end\n");
    /*输入0/1字符表示的二进制数，以#结束*/
    scanf("%c", &c);

    while(c != '#')
    {
        if(c == '0' || c == '1')
            Push(&s1, c);

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

    initStack(&s2);  /*创建一个栈s2，用来存放八进制字符串*/
    len = StackLen(s1);  /*得到栈中的元素个数，即二进制数的长度*/

    for(i = 0; i < len; i = i + 3) {
        for(j = 0; j < 3; j++) {
            Pop(&s1, &c); /*取出栈顶元素*/
            sum = sum + (c - 48) * pow(2, j); /*转换为八进制数*/

            if(s1.base == s1.top) break;
        }

        Push(&s2, sum + 48) ; /*将八进制数以字符形式压入栈中*/
        sum = 0;

    }

    printf("The Octal from is \n") ;

    while(s2.base != s2.top ) { /*输出八进制栈的内容*/
        Pop(&s2, &c);
        printf("%c", c);
    }
}