求助如何改时钟的两个--闪烁0.5秒程序如下

ID:79244 · 发表于 2015-5-19 23:36

#include<reg52.h>
  #define uchar unsigned char
  #define uint unsigned int
  #define play P0
  sbit LSA=P1^0;
  sbit LSB=P1^1;
  sbit LSC=P1^2;
  sbit led=P2^2; //闹钟检测
  sbit k1=P2^7;// 功能切换
  sbit k2=P2^6;//设置加
  sbit k3=P2^5;//  设置减
  sbit k4=P2^4; //设置位选
  sbit beet=P1^5; //蜂鸣器输出口
  uchar y;//按键松开
  uchar kai;//闹钟开启标志
  uchar she;//设置标志
  uchar t;//时间计数
  uchar num;//功能切换
  uchar fen;
  uchar shi;
  uchar miao;
  uchar yue;
  uchar nian;
  uchar ri;
  uchar shijian[8];
  uchar  shi1; //闹铃时
  uchar fen1;  //闹铃分
  unsigned char code shuma[12]={  0x3f,0x06,0x5b,0x4f,0x66,0x6d,0x7d,0x07,0x7f,0x6f,0x40,0x00};
  uchar brunyue[12]={31,28,31,30,31,30,31,31,30,31,30,31};
uchar runyue[12]={31,29,31,30,31,30,31,31,30,31,30,31};
void zhongshe(); void qishe();
void naoshe(); void shezhi();
void delay10ms(void) //误差 0us
{
unsigned char a,b,c;
for(c=1;c>0;c--)
for(b=38;b>0;b--)
for(a=130;a>0;a--);
}
void naozhong()
{
uint a,m;
if((shi1==shi)&&(fen1==fen)&&(kai%2))
{
for(m=20;m>0;m--)
for(a=10;a>0;a--)
if((k1==0)||(k2==0)||(k3==0)||(k4==0))
{ kai=0;break;
}

}
}
void plays()    //输出显示
{
unsigned char i;
unsigned int j;
for(i=0;i<8;i++)
{
switch(i)  //位选，选择点亮的数码管，
{
case(0):    LSA=0;LSB=0;LSC=0; break;//显示第0位
case(1):    LSA=1;LSB=0;LSC=0; break;//显示第1位
case(2):    LSA=0;LSB=1;LSC=0; break;//显示第2位
case(3):    LSA=1;LSB=1;LSC=0; break;//显示第3位
case(4):    LSA=0;LSB=0;LSC=1; break;//显示第4位
case(5):    LSA=1;LSB=0;LSC=1; break;//显示第5位
case(6):    LSA=0;LSB=1;LSC=1; break;//显示第6位
case(7):    LSA=1;LSB=1;LSC=1; break;//显示第7位
}
play=shuma[shijian[i]];
if(t>50)
{
switch(she)
{
case(3):
if((i==0)||(i==1))
play=0x00;
break;
   case(2):
if((i==3)||(i==4))
play=0x00; break;
case(1):
if((i==6)||(i==7))
play=0x00; break;
}
   }
   j=10;          //扫描间隔时间设定
   while(j--);
   play=0x00;//消隐
   }
   }
   void nao()    //闹钟赋值
   {
   shijian[7]=kai;
   shijian[6]=11;
   shijian[5]=10;
   shijian[4]=fen1%10;
   shijian[3]=fen1/10;
   shijian[2]=10;
   shijian[1]=shi1%10;
   shijian[0]=shi1/10;
   }
   void rili()       //日期赋值
   {
   shijian[7]=ri%10;
   shijian[6]=ri/10;
   shijian[5]=10;
   shijian[4]=yue%10;
   shijian[3]=yue/10;
   shijian[2]=10;
   shijian[1]=nian%10;
   shijian[0]=nian/10;
   }
   void zhong()       //电子钟赋值
   {
   shijian[7]=miao%10;
   shijian[6]=miao/10;
   shijian[5]=10;
   shijian[4]=fen%10;
   shijian[3]=fen/10;
   shijian[2]=10;
   shijian[1]=shi%10;
   shijian[0]=shi/10;

   }
   void InitTimer0(void)
   {
   TMOD = 0x01;
   TH0 = 0x0D8;
   TL0 = 0x0F0;
   EA = 1;
   ET0 = 1;
   TR0 = 1;
      }
void shezhi()       //功能设置函数
{
if((she!=0)&&((k2==0)||(k3==0)))
{
delay10ms();
if((she!=0)&&((k2==0)||(k3==0)))
{
switch(num)
{
case(0):
   zhongshe(); break;
   case(1):
   qishe(); break;
   case(2):
   naoshe();break;
   }
   }
   }
   }
   void main()
   {


   miao=0;fen=0; shi=0;
   ri=19;  yue=5; nian=15;
   InitTimer0();
   while(1)
   {
   if(k1==0)
   {
   delay10ms();
   if(k1==0)
   {num=(num+4)%3;

   }

      }
if(k4==0)
{
delay10ms();
   if(k4==0)
   she=(she+5)%4;
   while(!k4);
   }
   shezhi();

   plays();
   }
   }
   void Timer0Interrupt(void) interrupt 1
   {
   TH0 = 0x0D8;
   TL0 = 0x0F0;
      t++;
if(t==100)
{
t=0;
miao++;
if(miao>59)
{
miao=0;
fen++;
if(fen>59)
{
fen=0;
shi++;
if(shi>23)
{
shi=0;
   ri++;
   if(nian%4==0)
   {
   if(ri>runyue[yue-1])
   {
   ri=1;
   yue++;
   }
   if(yue>12)
   {
   yue=1;
   nian++;
   }
   }
   else
   {
   if(ri>brunyue[yue-1])
   {
   ri=0;
   yue++;
      }
if(yue>12)
{
yue=0;

               nian++ ;

   }

   }
               }
            }
      }}

switch(num)
{
   case(0):    zhong();break;
   case(1):    rili();break;

   default: zhong();
   }
   }
   void zhongshe()
   {
   switch(she)
   {
   case(1):    miao=0;InitTimer0();break;
   case(2):    if(k2==0)    fen=(fen+61)%60;
   else             fen=(fen+59)%60; break;
   case(3):  if(k2==0)    shi=(shi+25)%24;
   else       shi=(shi+23)%24;
   break;
   }
   while((!k2)||(!k3));
   }
   void naoshe()
      {
   switch(she)
   {
   case(1):    if(kai==1) kai=0;
   else kai=1; break;
   case(2):    if(k2==0)    fen1=(fen1+61)%60;
   else             fen1=(fen1+59)%60; break;
   case(3):  if(k2==0)    shi1=(shi1+25)%24;
   else       shi1=(shi1+23)%24;       break;
   }
   while((!k2)||(!k3));
   }
   void qishe()
   {
   switch(she)
   {
   case(1):
      if(k2==0)
{if(nian%4==0)
   ri=(ri+runyue[yue-1]+2)%(runyue[yue-1]+1) ;
   else
   ri=(ri+brunyue[yue-1]+2)%(brunyue[yue-1]+1);
   }
   else
   {
   if(nian%4==0) ri=(ri+runyue[yue-1])%(runyue[yue-1]+1) ;
   else
   ri=(ri+brunyue[yue-1])%(brunyue[yue-1]+1);
   }
      break;
case(2):  if(k2==0)    yue=(yue+14)%13;
else    yue=(yue+12)%13;
break;
case(3):  if(k2==0)    nian++;
else    nian--;
}
while((!k2)||(!k3));
}