單獨超聲波測距的時候好使，單獨舵機轉(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
} |
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