数据结构--二进制转化为十进制的实现

这个学期迫于学习数据结构，没办法，把书上的例子重新敲了一遍。

使用栈实现二进制转化为十进制

   #include <stdio.h>
#include <stdlib.h>
#include <math.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)
{
s->base = (ElemType *)malloc(STACK_INIT_SIZE*(sizeof(ElemType)));
if(!s->base)
{
exit(0);
}
s->top = s->base;
s->stackSize = STACK_INIT_SIZE;
}
void Push(sqStack *s,ElemType e)
{
if(s->top-s->base >= s->stackSize)
{
//内存不够
s->base = (ElemType *)realloc(s->base,(s->stackSize+STACKINCREMENT)*sizeof(ElemType));
if(!s->base){
exit(0);
}
}
*(s->top) = e;
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 s;
int len,i,sum = 0;
InitStack(&s);
printf("请输入二进制数，以#符号表示结束\n");
scanf("%c",&c);
while(c != '#')
{
Push(&s,c);
scanf("%c",&c);
}
getchar(); //把回车从缓冲区去掉，
len = StackLen(s);
printf("栈的当前容量是%d\n",len);
for(i=0;i<len;i++)
{
Pop(&s,&c);
sum = sum +(c-48)*pow(8,i);
}
printf("转化为八进制数为%d",sum);
return 0;
}