51单片机加法器按键清屏再恢复求解 -

标题: 51单片机加法器按键清屏再恢复求解 [打印本页]

作者: pov001 时间: 2021-2-12 23:51
标题: 51单片机加法器按键清屏再恢复求解

这是一个加法器，一个5道题，答对一题亮1盏灯，需要在按键增加其他功能，目前我完成了一部分功能，我就想请教那个按一下清屏，再按一下恢复那个怎么实现，程序是什么。。。。

程序如下图
#include<reg51.h>
unsigned char guan[13]={0x3f,0x06,0x5b,0x4f,0x66,0x6d,0x7d,0x07,0x7f,0x6f,0x77,0x48,0x00};
unsigned char shu[8]={0xf8,0xf9,0xfa,0xfb,0xfc,0xfd,0xfe,0xff};
unsigned char ggboy[6]={0xff,0xf7,0xe7,0xc7,0x87,0x07};
unsigned char hang[4]={0x08,0x04,0x02,0x01};
unsigned char lie[4]={0xef,0xdf,0xbf,0x7f};
unsigned char key;
int seg1[6]={1,2,3,4,5,6};
int seg2;
int sum;int n=0,p=9,t=0;
int c=0,d=0,v=0,z=0,num=0;
int n13=1;
void delay(int a)
{int j;
j=a;while(j--);}

void df() interrupt 1 //计时功能
{ int h;
      TH0=(65536-50000)/256;
TL0=(65536-50000)%256;
n=n+1;
      if(n==100)
      {for(h=0;h<=4;h++)
            {P0=shu[6]&ggboy[5];
                     P1=guan[seg1[d]];
                     delay(10000000);
                     P0=shu[7]&ggboy[0];
                     P1=0X00;
   delay(10000000);}n=0;}}

void sdwoig() //换题按钮
{ if(v==1)
      {d=d+1;
      if(d==6)
      {d=0;}}
      if(v==2)
      {d=d-1;
  if(d==-1)
  {d=5;}}
      if(v==3)
      {d=d-1;z=z-1;
      }}





void chengjichaxun()       //查成绩功能
{int j=1000;
      while(j--)
      {P0=(shu[0]&ggboy[z]);
P1=0x00;
      delay(1000);
      P0=(shu[1]&ggboy[z]);
P1=0x5e;
      delay(1000);
      P0=(shu[2]&ggboy[z]);
      P1=guan[z];
            delay(1000);
      P0=(shu[3]&ggboy[z]);
      P1=0x40;
            delay(1000);
      P0=(shu[4]&ggboy[z]);
      P1=0x39;
      delay(1000);
P0=shu[5]&ggboy[z];
P1=guan[(5-z)];
            delay(1000);
P0=shu[6]&ggboy[z];
P1=0x40;
            delay(1000);
      P0=shu[7]&ggboy[z];
P1=0x00;
         delay(1000);
         }

}

void sg1()       //数码管显示内容
{
      P0=(shu[1]&ggboy[z]);
P1=guan[seg1[d]];
      delay(1000);
      P0=(shu[2]&ggboy[z]);
      P1=guan[10];
            delay(1000);
      P0=(shu[3]&ggboy[z]);
            seg2=seg1[d]-1;
   sum=seg1[d]+seg2;
      P1=guan[seg2];
            delay(1000);
      P0=(shu[4]&ggboy[z]);
      P1=guan[11];
      delay(1000);
P0=shu[5]&ggboy[z];
P1=guan[num];
   delay(1000);    }

void ig()       //按键扫描与键盘功能
{ int a,t;
      P2=0x0f;
      if(P2!=0x0f)
      { delay(10000);
      if(P2!=0x0f)
      {
  for(a=0;a<=3;a++)
      {P2=lie[a];
      for(t=0;t<=3;t++)
      {if((P2&hang[t])==0)
  { key=a+4*t;}}}
      switch(key)
      { case 0:num=9;break;
   case 1:num=8;break;
               case 2:num=6;break;
               case 3:num=3;break;
               case 4:num=7;break;
               case 5:num=5;break;
               case 6:num=2;break;
               case 7:v=2;num=12;break;    //上一题
               case 8:num=4;break;
               case 9:num=1;break;
               case 10:                   //功能一彩灯功能
               case 11:v=1;num=12;break;    //下一题
               case 12:num=0;break;
               case 13:                      //功能二关机重启
               case 14:chengjichaxun();break; //功能三成绩查询
   case 15: //重做功能

if(num==sum)
{z=z+1;
if(z==6)
{z=1;}P0=(shu[5]&ggboy[z]);}
   delay(1000);

if(num<=9)
{v=1;}
if(v!=0)
{sdwoig();}
}}}}

void main()
{          TMOD=0x11;
                     TH0=(65536-50000)/256;
                     TL0=(65536-50000)%256;


         TR0=1;


         IE=0x9f;
      while(1)
      {sg1();delay(1000);{ig();delay(100);}}}

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