舵機(jī)轉(zhuǎn)動超聲波避障程序的問題

ID:85492 · 發(fā)表于 2015-10-19 16:53

單獨(dú)超聲波測距的時候好使，單獨(dú)舵機(jī)轉(zhuǎn)動的時候也好使，但是把這兩個程序組合到一起的時候就不好使了，求各位兄弟姐妹幫我看看程序是哪里出錯了。
#include"stc12c5a60s2.h"
#define uint unsigned int
#define uchar unsigned char
sbit trig=P0^2;
sbit echo=P3^2;
sbit led0=P0^0;
sbit pwm=P2^5;
bit flag;
float distance;
uchar count;
uchar jd;
uint time,timeH,timeL;
void delayms(uint x)
{
      uint i,j;
      for(i=x;i>0;i--)
            for(j=110;j>0;j--);
}
void delay20us()
{
      uint i;
      for(i=0;i<4;i++);
}
void duoji(uchar x)
{
      jd=x;
      count=0;
EA=1;
      ET1=1;
      TH1=0xfe;
TL1=0x33; //0.5ms
TR1=1;
}
void main()
{
      TMOD=0x11;
      led0=0;
      while(1)
      {
            EA=0;
            trig=1;
            delay20us();
            trig=0;
            while(echo==0);
            flag=0;
            EA=1;
            EX0=1;
            TH0=0;
            TL0=0;
            TF0=0;
            TR0=1;
            delayms(20);
            TR1=0;
            if(flag==1)
            {
                     time=timeH*256+timeL;
                     distance=1.085*time*0.017;
                     if(distance<10)
                     {
                              duoji(2);
                              delayms(5000);
                              duoji(3);
                              delayms(5000);
                              duoji(4);
                              delayms(5000);
                              duoji(3);
                     }
                     else
                     {
                              led0=1;
                     }
            }
            if(flag==0)
            {
                     distance=0;
            }
      }
}
void exter() interrupt 0
{
      timeH=TH0;
      timeL=TL0;
      flag=1;
      EX0=0;
}
void t0() interrupt 1
{
      TH0=0;
      TL0=0;
}
void t1() interrupt 3
{
      TH1=0xfe;
      TL1=0x33;
      if(count<jd) //判斷0.5ms次數(shù)是否小于角度標(biāo)識
            pwm=1; //確實(shí)小于，PWM輸出高電平
      else
            pwm=0; //大于則輸出低電平
         count=(count+1); //0.5ms次數(shù)加1
         count=count%40; //次數(shù)始終保持為40 即保持周期為20m
}