逆波兰表达式c算法

*示例：输入 1 2 - 4 5 + ，算出它的逆波兰表达式值。
代码解决：
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>//这是啥

#define STACK_INIT_SIZE 20
#define STACKINCREMENT 10
#define MAXBUFFER 10

typedef double ElemType;
typedef struct
{
ElemType *base;
ElemType *top;
int stackSize;
}sqStack;

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;
}
Push(sqStack *s,ElemType e)
{
if(s->top - s->base >=s->stackSize)
{
//栈空间不够用,用realloc函数增加空间
s->base = (ElemType *)realloc(s->base,(s->stackSize + STACKINCREMENT)*sizeof(ElemType));
if(!s->base)
{
exit(0);
}
}
*(s->top) = e;
s->top++;
}
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()
{
sqStack s;
char c;
double d,e;
char str[MAXBUFFER];

int i = 0;
InitStack(&s);
printf("请按逆波兰表达式输入待计算数据，数据与运算符之间按空格隔开,以#作为结束标志：");
scanf("%c",&c);
while(c != '#')
{
    while(isdigit(c)||c == '.')//用于过滤数字
    {
        str[i++] = c;
        str[i] = '\0';
        if(i>=10)
        {
            printf("出错，输入的单个数据过大!\n");
            return -1;
        }
        scanf("%c",&c);
        if( c == ' ')
        {
            d = atof(str);//将字符串转换为float
            Push(&s,d);
            i = 0;
            break;
        }
    }
    switch(c)
    {
        case '+':
            Pop(&s,&e);
            Pop(&s,&d);
            Push(&s,d+e);
            break;
        case '-':
           Pop(&s,&e);
           Pop(&s,&d);
           Push(&s,d-e);
           break;
        case '*':
           Pop(&s,&e);
           Pop(&s,&d);
           Push(&s,d*e);
           break;
        case '/':
           Pop(&s,&e);
           Pop(&s,&d);
           if(e != 0)
           {
             Push(&s,d/e);
           }else{
            printf("\n出错：除数为0！\n");
            return -1;
           }
           break;
    }
    scanf("%c",&c);//这个作用是不断输入字符，因为每输一个scanf函数就使用完了
}
Pop(&s,&d);
printf("最终的计算结果为：%d\n",d);
return 0;

}