发现AD转换部分只转换一次便不动

只看该作者 · 2020-3-5 18:53

在AD转换时没出现问题，后来要求加个串口，于是我加上了串口程序，一仿真，发现AD转换部分只转换一次便不动了。。

uchar disbuf[]={'0','0','0',0x00};
sbit rs=P2^0;
sbit en=P2^1;
sbit rw=P2^2;
sbit OE=P2^3;
sbit EOC=P2^4;
sbit ST=P2^5;
sbit CLK=P2^6;

void delay(uint z)
{
uint x,y;
      for(x=z;x>0;x--)
            for(y=110;y>0;y--);
}

void command(uchar com)
{
      rs=0;
      delay(5);
      P1=com;
      en=1;
      delay(5);
      en=0;
}

void figure(uchar figure)
{
      rs=1;
      delay(5);
      P1=figure;
      en=1;
      delay(5);
      en=0;
}

void write_first(uchar *s)
{
      command(0x80);
      while(*s)
      {
            figure(*s);
            s++;
            delay(20);
      }
}

void write_second(uchar x)
{
      command(0xc0);
      disbuf[0]=x/100+0x30;
      disbuf[1]=x%100/10+0x30;
      disbuf[2]=x%10+0x30;
      for(i=0;i<3;i++)
      {
               figure(disbuf);
               delay(5);
      }
}

void init()
{
      rw=0;
      TMOD=0x22;
      SCON=0X50;
      PCON=0;
      TH1=TL1=0XFD;
      TH0=(256-4)/256;
      TL0=(256-4)%256;
      command(0x38);
      command(0x0c);
      command(0x06);
      command(0x80);
}

void main()
{
      uchar temp,i;
      uchar *p;
      init();
      p="ADconvert result";;
      write_first(p);
      EA=1;
      ET0=1;
      TR0=1;
      TR1=1;
      while(1)
      {
               ST=0;ST=1;ST=0;
               delay(5);
               while(EOC==0);
               OE=1;
               temp=P0;
               write_second(temp);
               OE=0;
               i=0;
               while(disbuf!=0x00)
               {
                     SBUF=disbuf;
                     while(TI==0);
                     TI=0;
                     i++;
                     delay(5);
               }
               while(RI==0);
               RI=0;
               while(TI==0);
               TI=0;
      }
}

void timer0() interrupt 1
{
   CLK=~CLK;
}