4位数码管扫描程序

1万 · 2009-5-24 20:20

void  display(void)
{  uint i,j;
   j=0xfe;
  for(i=0;i<4;i++)
  {
   P2=shu[tm[3-i]];
   P0=~(0x01<<i)|0x80;

   delay(2);
  }


  }

为何最后一位比其它的位亮？

1万 · 2009-5-24 20:22

1万 · 2009-5-24 20:59

Display$Temp:PROCEDURE PUB;
    Temp = Temperature / 16;
    Disp$Buf(0) = Char$Img(Temp MOD 10);
    Disp$Buf(1) = Char$Img((Temp / 10) MOD 10);
END Display$Temp;

System$Clock:PROCEDURE USING 1;
/* 1ms 一次 */
    MSecond$Cnt = MSecond$Cnt + 1;

    IF MSecond$Cnt MOD 4 = 0 THEN
    DO;
        LED1 = 1;
        LED2 = 1;
        P1 = Disp$Buf(Disp$Buf$Ptr);
        IF Disp$Buf$Ptr = 0 THEN
            LED1 = 0;
        ELSE IF Disp$Buf$Ptr = 1 THEN
            LED2 = 0;

        Disp$Buf$Ptr = Disp$Buf$Ptr + 1;
        IF Disp$Buf$Ptr >= 2 THEN
            Disp$Buf$Ptr = 0;
    END;

    IF MSecond$Cnt < 1000 THEN
        RETURN;

    MSecond$Cnt = 0;
    Second = Second + 1;

    IF FanOutDelay > 0 THEN
        FanOutDelay = FanOutDelay - 1;

END System$Clock;

/*=========================================================================*/
clk$Inter: PROCEDURE INTERRUPT 1 USING 1;
    TR0 = 0;
    TH0 = HIGH(-Timer0Cnt + 34);
    TL0 = LOW(-Timer0Cnt + 34);
    TR0 = 1;

    CALL System$Clock;
END clk$Inter;