基于逆波兰表达式的计算器

基于逆波兰表达式的计算器

首先我们需要了解下一般表达式如何转化为逆波兰表达式。如下:

1+(20-3)*4+10/5 如何转化为逆波兰表达式 1 20 3 - 4 * + 10 5 / +

一般表达式转化为逆波兰表达式的步骤如下：

从左到右遍历一般表达式的每个数字和符号，若是数字则直接输出，若是符号，则判断其与栈顶符号的优先级，是右括号或者优先级低于栈顶符号，则栈顶元素依次出栈并输出，直到遇到左括号或栈空才将等待入栈的符号入栈。

代码如下

#include<time.h>
#include<math.h>
#include<ctype.h>
# include <stdio.h>
# include <stdlib.h>
#include<string.h>

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

typedef char ElemType;

typedef struct
{

       ElemType *base;//栈底指针
       ElemType *top;//栈顶指针
      int stacksize;//栈当前可使用的最大容量
}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)
    {
        s->base = ( ElemType*)realloc(s->base,(STACKINCREMENT + s->stacksize)*sizeof(ElemType));
        if(!(s->base))exit(0);
        s->top = s->base+ s->stacksize;
        s->stacksize = s->stacksize + STACKINCREMENT;
    }
    *(s->top) = e;
    (s->top)++;
}
void Pob(sqstack *s, ElemType *e)
{
    if(s->top==s->base)return ;
    s->top--;
    *e = *(s->top);
}

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

typedef double ElemType1;

typedef struct
{

       ElemType1 *base;//栈底指针
       ElemType1 *top;//栈顶指针
      int stacksize;//栈当前可使用的最大容量
}sqstack1;

void InitStack1(sqstack1 *s)
{
    s->base = ( ElemType1*)malloc(STACK_INIT_SIZE*sizeof( ElemType1));
    if(!(s->base))exit(0);
    s->top = s->base;
    s->stacksize = STACK_INIT_SIZE;
}
void Push1(sqstack1  *s, ElemType1 e)
{
    if(s->top - s->base >= s->stacksize)
    {
        s->base = ( ElemType1*)realloc(s->base,(STACKINCREMENT + s->stacksize)*sizeof(ElemType1));
        if(!(s->base))exit(0);
        s->top = s->base+ s->stacksize;
        s->stacksize = s->stacksize + STACKINCREMENT;
    }
    *(s->top) = e;
    (s->top)++;
}
void Pob1(sqstack1 *s, ElemType1 *e)
{
    if(s->top==s->base)return ;
    s->top--;
    *e = *(s->top);
}

int StackLen1(sqstack1 s)
{
    return (s.top - s.base);
}

void midConverToLast(char * &str,int n)
{
	str = (char*)malloc(n*sizeof(char));
	
	sqstack s;
	char c,e;
	InitStack(&s);
	printf("请输入中缀表达式,以#作为结束标志：");
	scanf("%c",&c);


	int i = 0;
	while('#'!=c)
	{

		while(c >= '0' && c <= '9')
		{
			//printf("%c",c);
			str[i++] = c;
			scanf("%c",&c);
			if(c < '0' || c > '9')
			{
					//printf(" ");
				str[i++] = ' ';
			}
		}
		if(c == ')')
		{
			Pob(&s,&e);
			while(e != '(')
			{
				//printf("%c ",e);
				str[i++] = e;
				str[i++] = ' ';
				Pob(&s,&e);

			}
			
		}
		else if(c =='+'||c=='-')
		{
			if(!StackLen(s))
			{
				Push(&s,c);
			}
			else
			{
				do
				{
					Pob(&s,&e);
					if('('==e)
					{
						Push(&s,e);
					}
					else
					{
						//printf("%c ",e);
						str[i++] = e;
				       str[i++] = ' ';
					}
				}while(StackLen(s) && e!='(');

				Push(&s,c);

			}

		}
		else if(c=='*'||c=='/'||'('==c)
		{
			Push(&s,c);
		}
		else if(c=='#')
		{
			break;
		}
		else 
		{
					printf("%n出错：输入格式错误\n ");
					
		}

			scanf("%c",&c);
	}
	while(StackLen(s))
	{
		Pob(&s,&e);
		str[i++] = e;
		str[i++] = ' ';
	}
	str[i]='\0';

}

int main()
{
	sqstack1 s;

	char c;
	int i = 0;
	double d,e;
	InitStack1(&s);
	
	char str[MAXBUFFER];
	char *point = NULL;
	int n = 30;
	midConverToLast(point,n);
	int j = 0;
	while((c = point[j])!='\0')
	{
		
		while(isdigit(c) || c=='.')
		{
			str[i++] = c;
			str[i] ='\0';
			if(i>=10)
			{
				printf("输入的单个数据位数过长\n");
				return -1;
			}
			c = point[++j];
			if(c==' ')
			{
				d = atof(str);
				Push1(&s,d);
				i = 0;
				break;
			}

		}
		switch(c)
		{
		case '+':
			Pob1(&s,&e);
			Pob1(&s,&d);
			Push1(&s,d + e);
			printf("加法的结果为%f\n",d + e);
			break;
		case '-':
			Pob1(&s,&e);
			Pob1(&s,&d);
			Push1(&s,d - e);
			printf("减法的结果为%f\n",d - e);
			break;
		case '*':
			Pob1(&s,&e);
			Pob1(&s,&d);
			Push1(&s,d * e);
			printf("乘法的结果为%f\n",d * e);
			break;
		case '/':
			Pob1(&s,&e);
			Pob1(&s,&d);
			if(e!= 0)
			{
				Push1(&s,d/e);
			}
			else
			{
				printf("出错，除数为0\n");
				return -1;
			}
			printf("除法的结果为%f\n",d/e);
			break;
		}
	   c = point[++j];

	}
	Pob1(&s,&d);
	printf("最后的运行结果为%f\n",d);

	return 0;}