Proteus仿真具有清零功能秒表相关 -

标题: Proteus仿真具有清零功能秒表相关 [打印本页]

作者: 田所浩二单推人 时间: 2021-6-17 11:19
标题: Proteus仿真具有清零功能秒表相关
仿真图如下不知道为什么没法+1和清零
代码如下
#include<reg51.h>
#include<intrins.h>
#define uchar unsigned char
sbit K9=P2^4;
sbit K10=P2^5;
sbit K11=P2^6;
uchar code segtab[]=
{0xc0,0xf9,0xa4,0xb0,0x99,0x92,0x82,0xf8,0x80,
0x90,0x88,0x83,0xc6,0xa1,0x86,0x8e,0xff,0xbf};
uchar dbuf[4]={0,0,0,0x10};

void delay(void)
{ uchar i;
for(i=0;i<200;i++);
}
void disp(void)
{  uchar n,bsel;
bsel=0x01;
for(n=0;n<4;n++)
{ P0=segtab[dbuf[n]];
   P2=bsel|0xf0;
   bsel=_crol_(bsel,1);
   delay();
   P0=0xff;
}
}
void main(void)
{ uchar i;
int count=0;
unsigned int t;
while(1)
{disp();
if((P2&0x30)!=0x30)
{  for(i=0;i<10;i++)disp();
      if((P2&0x30)!=0x30)
      { if(K9==0)
         {  count++;
            if(count>999)count=-999;
         }
   if(K10==0)
   { count--;
   if(count<-999)count=999;
   }
   else
   count=0;
   if(count>=0)
   { dbuf[3]=0x10;
         t=count;
   }
   else
   { dbuf[3]=0x11;
         t=-count;
   }
   dbuf[2]=t/100;
   dbuf[1]=(t%100)/10;
   dbuf[0]=t%10;
   while((P2&0x30)!=0x30)disp();
}
}
}
}

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