点阵式.txt

简单的程序

#include <AT89X52.H>
unsigned char code tab[]={0xfe,0xfd,0xfb,0xf7,0xef,0xdf,0xbf,0x7f};
unsigned char code digittab[10][8]={	　 {0x00,0x00,0x3e,0x41,0x41,0x41,0x3e,0x00},	//0
                                    {0x00,0x00,0x00,0x00,0x21,0x7f,0x01,0x00},	//1
                                    {0x00,0x00,0x27,0x45,0x45,0x45,0x39,0x00},	//2
                                    {0x00,0x00,0x22,0x49,0x49,0x49,0x36,0x00},	//3
                                    {0x00,0x00,0x0c,0x14,0x24,0x7f,0x04,0x00},	//4
                                    {0x00,0x00,0x72,0x51,0x51,0x51,0x4e,0x00},	//5
                                    {0x00,0x00,0x3e,0x49,0x49,0x49,0x26,0x00},	//6
                                    {0x00,0x00,0x40,0x40,0x40,0x4f,0x70,0x00},	//7
                                    {0x00,0x00,0x36,0x49,0x49,0x49,0x36,0x00},	//8
                                    {0x00,0x00,0x32,0x49,0x49,0x49,0x3e,0x00}	//9
                                    };

unsigned int timecount;
unsigned char cnta;
unsigned char cntb;

void main(void)
{
  TMOD=0x01;
  TH0=(65536-3000)/256;
  TL0=(65536-3000)%256;
  TR0=1;
  ET0=1;
  EA=1;
  while(1)
    {;
    }
}

void t0(void) interrupt 1 using 0
{
  TH0=(65536-3000)/256;
  TL0=(65536-3000)%256;
  P3=tab[cnta];
  P1=digittab[cntb][cnta];
  cnta++;
  if(cnta==8)
    {
      cnta=0;
    }
  timecount++;
  if(timecount==333)
    {
      timecount=0;
      cntb++;
      if(cntb==10)
        {
          cntb=0;
        }
    }
}