51单片机T0定时器应用1

INT0X:
                                     MOV TH0,#(65536-50000) / 256
                                     MOV TL0,#(65536-50000) MOD 256
                                     INC TCOUNT
                                     MOV A,TCOUNT
                                     CJNE A,#20,NEXT
                                     MOV TCOUNT,#00H
                                     INC SECOND
                                     MOV A,SECOND
                                     CJNE A,#60,NEX
                                     MOV SECOND,#00H
NEX:                                     MOV A,SECOND
                                     MOV B,#10
                                     DIV AB
                                     MOV DPTR,#TABLE
                                     MOVC A,@A+DPTR
                                     MOV P0,A
                                     MOV A,B
                                     MOVC A,@A+DPTR
                                     MOV P2,A
NEXT:                          RETI
                  
TABLE:                        DB 3FH,06H,5BH,4FH,66H,6DH,7DH,07H,7FH,6FH
                                     END
9．C语言源程序（中断法）
#include
unsigned char code dispcode[]={0x3f,0x06,0x5b,0x4f,
                               0x66,0x6d,0x7d,0x07,
                               0x7f,0x6f,0x77,0x7c,
                               0x39,0x5e,0x79,0x71,0x00};
unsigned char second;
unsigned char tcount;
void main(void)
{
  TMOD=0x01;
  TH0=(65536-50000)/256;
  TL0=(65536-50000)%6;
  TR0=1;
  ET0=1;
  EA=1;
  tcount=0;
  second=0;
  P0=dispcode[second/10];
  P2=dispcode[second];
  while(1);
}
void t0(void) interrupt 1 using 0
{
  tcount++;
  if(tcount==20)
    {
      tcount=0;
      second++;
      if(second==60)
        {
          second=0;
        }
      P0=dispcode[second/10];
      P2=dispcode[second];             
    }
  TH0=(65536-50000)/256;
  TL0=(65536-50000)%6;
}

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