关于AT89c51单片机中用AT24C02的问题 -

标题: 关于AT89c51单片机中用AT24C02的问题 [打印本页]

作者: xuezha 时间: 2020-6-3 13:01
标题: 关于AT89c51单片机中用AT24C02的问题
代码怎么修改可以实现图上的内容有AT24C02的头文件

51hei截图20200603130003.png (115.9 KB, 下载次数: 48)

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IMG_20200603_125747.jpg (55.17 KB, 下载次数: 51)

作者: xuezha 时间: 2020-6-3 13:06
前面的程序图不完整C:\Users\19429\Desktop

作者: xuezha 时间: 2020-6-3 13:15
#include<reg51.h>
#include<AT24C02.h>
#define uchar unsigned char
#define       uint unsigned int
sbit P3_2=P3^2;
uchar code led_mod[]={0x3f,0x06,0x5b,0x4f,0x66,
0x6d,0x7d,0x07,0x7f,0x6f,0x00};
uchar code BitCode[]={0x7f,0xbf,0xdf,0xef,0xf7,0xfb,0xfd,0xfe};
uchar huanchong[]={10,10,10,10,10,10,10,10};
uchar convert[]={10,10,10,10,10,10,10,10};
uchar a=0;

void int0()  interrupt 0
{
      uchar code RowCode[]={0xfe,0xfd,0xfb,0xf7};
      uchar temp,row,column,i,y,z;
      uchar m,n;
      P1=0xf0;
      temp = P1 & 0xf0;
         if (temp != 0xf0)
                     {
   P2=0xff;
   switch (temp)
               {
                              case 0x70: column = 3; break;
                              case 0xb0: column = 2; break;
                              case 0xd0: column = 1; break;
                              case 0xe0: column = 0; break;
                              default: break;
                     }
                     for (i=0;i<4;i++)
                     {
                              P1 = RowCode[i];
                              temp = P1 & 0xf0;
                              if (temp !=0xf0)
                              {
                                       row = i;break;
                              }
                     }
                     y= row *4 + column;
                     if(y<10)
                     {
                              if(a<8)
                              {
                                       huanchong[a++]=y;
                           m=a;
                                       for(n=0;n<a;n++)
                     {
                              convert[n]=huanchong[--m];
                           }
                              }
                     }
         if(y>=10)
                     {
                              P2=0xff;
                              a=0;
                              for(z=0;z<8;z++)
                              {
                                       huanchong[z]=10;
                                       convert[z]=10;
                              }
               }
                     while((P1 & 0xf0) !=0xf0);
      }

}

void BufToSeg(void)
{
      uchar i=0;
      for(;i<8;i++)
      {

            P0=led_mod[10];
            P2= BitCode[i];
            P0=led_mod[convert[i]];
      }
}

void main(void)
{
      EA=1;
      EX0=1;
      IT0=1;
      while(1)
      {
            P1=0x0f;
            BufToSeg();
      }
}

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