基于单片机四则运算计算器设计+proteus仿真

• 设计要求
• 硬件设计
• 软件设计流程图
• 程序代码

设计要求

使用keil c软件基于单片机系统设计一个“简单的计算器”。要求设计的模拟计算器必须从按钮输入0~9组成一个数，再从按钮输入加、减、乘和除运算符（+、-、*、/），同样从按钮上输入另一个数，之后按下等于按钮能够实现两个数之间的和、差、积和商的运算并且显示相关的操作步骤。

硬件设计

1、AT89C51单片机
2、手动复位电路
3、时钟电路
4、LCD602液晶显示
5、4*4按键矩阵

软件设计流程图

程序代码

#include "reg52.h"
#include "stdio.h"
#include "key.h"
#include "1602.h"
#define uchar unsigned char
#define uint unsigned int
uchar Line1[20];  //第一行显示数组
uchar Line2[20];  //第二行显示数组
uint key_flag;     //按键标志
int result1 = 0; //存储第一个整数
int result2 = 0;
int Oper = 0;  //存储运算符
uint succeed_flag = 0;  //计算完成标志

void display_line1(uint key)
{
    static uint num = 0;
    uchar str = 0;
 
    if(succeed_flag == 1)
    {    
        sprintf((char *)Line1,"                 ");  //直接输入空格便会清除
        print_string(Line1,1);
        sprintf((char *)Line2,"                 ");
        print_string(Line2,2);
        succeed_flag = 0;
        num = 0;
    }    
    switch(key)
    {
        case 0 :  str = '1';  break;
        case 1 :  str = '2';  break;
        case 2 :  str = '3';  break;
        case 3 :  str = '+';  break;
        case 4 :  str = '4';  break;
        case 5 :  str = '5';  break;
        case 6 :  str = '6';  break;
        case 7 :  str = '-';  break;
        case 8 :  str = '7';  break;
        case 9 :  str = '8';  break;
        case 10 : str = '9';   break;
        case 11 : str = '*';   break;
        case 12 :              break;
        case 13 : str = '0';   break;
        case 14 : str = '=';   break;
        case 15 : str = '/';   break;
      default:break;
    }
    Line1[num] = str;
    print_string(Line1,1);
    num ++;
    if(num >= 20)
        num = 0;
}

void distinguish_num1(uint key)
{
    static int num = 0;
 
    switch(key)
    {
        case 0 :  num = num * 10 + 1;  break;
        case 1 :  num = num * 10 + 2;  break;
        case 2 :  num = num * 10 + 3;  break;
        case 4 :  num = num * 10 + 4;  break;
        case 5 :  num = num * 10 + 5;  break;
        case 6 :  num = num * 10 + 6;  break;
        case 8 :  num = num * 10 + 7;  break;
        case 9 :  num = num * 10 + 8;  break;
        case 10 : num = num * 10 + 9;   break;
        case 13 : num = num * 10 + 0;   break;
      default:break;
    }    
 
    if(key == 3)
    {
        Oper = 1;
        result1 = num;
        num = 0;   //将num清零，避免影响之后的计算
    }
    else if(key == 7)
    {
        Oper = 2;
        result1 = num;
        num = 0;   //将num清零，避免影响之后的计算
    }
    else if(key == 11)
    {
        Oper = 3;
        result1 = num;
        num = 0;    //将num清零，避免影响之后的计算
    }
    else if(key == 15)
    {
        Oper = 4;
        result1 = num;
        num = 0;    //将num清零，避免影响之后的计算
    }
}

void distinguish_num2(uint key)
{
    static int num = 0;
    switch(key)
    {
        case 0 :  num = num * 10 + 1;  break;
        case 1 :  num = num * 10 + 2;  break;
        case 2 :  num = num * 10 + 3;  break;
        case 4 :  num = num * 10 + 4;  break;
        case 5 :  num = num * 10 + 5;  break;
        case 6 :  num = num * 10 + 6;  break;
        case 8 :  num = num * 10 + 7;  break;
        case 9 :  num = num * 10 + 8;  break;
        case 10 : num = num * 10 + 9;   break;
        case 13 : num = num * 10 + 0;   break;
      default:break;
    }    
    if(key == 14)
    {
        result2 = num;
        if(Oper == 1)  
        {
            result2 = result1 + result2;            
        }
        else if(Oper == 2)  
        {
            result2 = result1 - result2;
        }
        else if(Oper == 3)  
        {
            result2 = result1 * result2;
        }
        else if(Oper == 4)  
        {
            result2 = result1 / result2;
        }        
        sprintf((char *)Line2,"%d   ",result2);
        print_string(Line2,2);
        result1 = 0;
        result2 = 0;
        Oper = 0;
        num = 0;
        succeed_flag = 1;
    }
}
void main()
{  
     lcd_init();   //lcd初始化
   while(1)
   {
       key_flag = key_scan();   //按键检测     没有按键被按下时 返回值为16
        if(key_flag != 16)    //有按键被按下
        {
                display_line1(key_flag);
              if(Oper == 0)
                {
                    distinguish_num1(key_flag);
                }
                else
                {
                    distinguish_num2(key_flag);                    
                }
        }
   }
}

程序设计图：
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