数据结构小甲鱼逆波兰计算器学习笔记

问题：（1-2）*（4-5） 逆波兰表达式输出为 1 2 - 4 5 + *

#include
#include
#include 

#define STACK_INIT_SIZE 20
#define STACKICREAMENT  10
#define MAXBUFFER 10

typedef double ElemType;
typedef struct
{
	ElemType *top;
	ElemType *base;
	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)
	{
		s->base = (ElemType *)realloc(s->base, (s->stacksize *STACKICREAMENT)*sizeof(ElemType));
		if (!s->base)
		{
			exit(0);
		}
		s->top = s->base + s->stacksize;
		s->stacksize = s->stacksize + STACKICREAMENT;
	}
	*(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(void)
{
	sqStack s;
	char c;
	double d, e;

	char str[MAXBUFFER];//缓冲区 
	int i = 0;
	InitStack(&s);
	printf("请按逆波兰表达式输入待计算数据，数据与运算符之间用空格隔开,以#作为结束标志：");
	scanf("%c", &c);

	while (c != '#')
	{
		while (isdigit(c) || c == '.')//过滤数字操作
		{
			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);//atof字符串到double的转换  
				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：除数为零！\n");
				return -1;
			}
			break;

		}
		scanf("%c", &c);
	}
	Pop(&s, &d);

	printf("\n最终计算结果为：%f\n", d);
	system("pause");
	return 0;
}