C语言实现一个简单的计算器

来源：互联网发布：全国省市区数据库编辑：程序博客网时间：2024/04/25 16:38

#include<stdio.h>
#include<stdlib.h>
#define N 100

char stack1[N];
char stack2[N];
int top = -1;

char pop(char *stack)
{
    char tmp;
    if(top == -1)
    {
        printf("该栈为空！\n");
    }
    else
    {
        tmp = stack[top];
top--;
    }
    return tmp;
}
void push(char *stack,char data)
{
    if(top == N-1)
    {
        printf("该栈已满！\n");
    }
    else
    {
        top++;
stack[top] = data;
    }
}

char *topostfix(char *expstr)
{
    char *postfix = (char *)malloc(sizeof(char));
    int i = 0;
    int j = 0;
    int tmp;
    while(expstr[i] != '\0')
    {
        switch(expstr[i])
{
     case '+':
     case '-':
         while(top != -1 && stack1[top] != '(')
  {
             postfix[j++] = pop(stack1);
  }
  push(stack1,expstr[i++]);
  break;
     case '*':
     case '/':
         while(top != -1 && (stack1[top] == '*' || stack1[top] == '/'))
  {
      postfix[j++] = pop(stack1);
  }
  push(stack1,expstr[i++]);
  break;
     case '(':
         push(stack1,expstr[i++]);
  break;
     case ')':
         tmp = pop(stack1);
  while(top != -1 && tmp != '(')
  {
      postfix[j++] = tmp;
      tmp = pop(stack1);
  }
  i++;
         break;
     default :
         while(expstr[i] >= '0' && expstr[i] <= '9'
                                && expstr[i] != '\0')
         {
             postfix[j++] = expstr[i++];
         }
  postfix[j++] = ' ';
  break;
}
    }
    while(top != -1)
    {
        postfix[j++] = pop(stack1);
    }
    postfix[j] = '\0';
    return postfix;
}
int operation(char *postfix)
{
    int i = 0;
    int result;
    while(postfix[i] != '\0')
    {
        if(postfix[i] >= '0' && postfix[i] <= '9')
{
     result = 0;
     while(postfix[i] != ' ')
     {
         result = result * 10 + (postfix[i] - '0');
  i++;
     }
     i++;
     push(stack2,result);
}
else
{
     if(postfix[i] != ' ')
     {
         int y = pop(stack2);
  int x = pop(stack2);
  switch(postfix[i])
  {
      case '+':
          result = x + y;
          break;
      case '-':
          result = x - y;
   break;
      case '*':
          result = x * y;
   break;
      case '/':
          result = x / y;
   break;
  }
  push(stack2,result);
     }
     i++;
}
    }
    return result;
}
int main()
{
    int result;
    char expstr[N];
    char *postfix = (char *)malloc(sizeof(char));
    printf("请输入表达式：");
    gets(expstr);
    printf("你输入的表达式为：");
    puts(expstr);
    postfix = topostfix(expstr);
    printf("你输入的表达式的后缀表达式为：");
    puts(postfix);
    result = operation(postfix);
    printf("%s=%d\n",expstr,result);
    return 0;
}
本来想实现浮点型的，myatof是写好了，但还有很多困难~~~~遂不大想写了~~
这个计算器只有+,-,*,/,()~~~~
给点点你们改的空间，将结果能输出为浮点型~~~~~
提示：强制类型转换~~~~