基于51单片机的智能红外循迹小车程序 -

标题: 基于51单片机的智能红外循迹小车程序 [打印本页]

作者: cqupt.qwert 时间: 2019-6-14 19:17
标题: 基于51单片机的智能红外循迹小车程序
最近刚结束单片机讲座的课，在写关于智能汽车的文章的同时自己做了一个简单的智能循迹小车，这里采用的是光电循迹模块，使用比较简单，
红外对管输出的电压模拟信号通过比较器转换为了数字电平，只需要根据输出的电平高低控制电机的正反转，这边附上部分单片机代码

#include "reg52.h"

typedef unsigned int u16;
typedef unsigned char u8;

sbit left1 = P0^0;
sbit left2 = P0^1;
sbit right1 = P0^2;
sbit right2 = P0^3;

sbit IN1 = P1^4;
sbit IN2 = P1^3;
sbit IN3 = P1^2;
sbit IN4 = P1^1;

void xunji();

void main()
{
while(1)
{
      xunji();
}
}

void TurnRight()
{
IN1 = 0;
IN2 = 1;

IN3 = 0;
IN4 = 1;
}

void TurnLeft()
{
IN1 = 1;
IN2 = 0;

IN3 = 1;
IN4 = 0;
}

void Forward()
{
IN1 = 1;
IN2 = 0;

IN3 = 0;
IN4 = 1;
}

void Stop()
{
IN1 = 0;
IN2 = 0;

IN3 = 0;
IN4 = 0;
}

void xunji()
{
unsigned char flag = 0;
if((left1 == 0)&&(left2 == 0)&&(right1 == 0)&&(right2 == 0)) //0 0 0 0
      flag = 0;

if((left1 == 0)&&(left2 == 0)&&(right1 == 0)&&(right2 == 1)) //0 0 0 1
      flag = 1;

if((left1 == 0)&&(left2 == 0)&&(right1 == 1)&&(right2 == 0)) //0 0 1 0
      flag = 0;

if((left1 == 0)&&(left2 == 0)&&(right1 == 1)&&(right2 == 1)) //0 0 1 1
      flag = 1;

if((left1 == 0)&&(left2 == 1)&&(right1 == 0)&&(right2 == 0)) //0 1 0 0
      flag = 0;

if((left1 == 0)&&(left2 == 1)&&(right1 == 0)&&(right2 == 1)) //0 1 0 1
flag = 1;

if((left1 == 0)&&(left2 == 1)&&(right1 == 1)&&(right2 == 0)) //0 1 1 0
      flag = 0;

if((left1 == 0)&&(left2 == 1)&&(right1 == 1)&&(right2 == 1)) //0 1 1 1
      flag = 1;

if((left1 == 1)&&(left2 == 0)&&(right1 == 0)&&(right2 == 0)) //1 0 0 0
      flag = 2;

if((left1 == 1)&&(left2 == 0)&&(right1 == 0)&&(right2 == 1)) //1 0 0 1
      flag = 0;

if((left1 == 1)&&(left2 == 0)&&(right1 == 1)&&(right2 == 0)) //1 0 1 0
  flag = 2;

if((left1 == 1)&&(left2 == 0)&&(right1 == 1)&&(right2 == 1)) //1 0 1 1
      flag = 0;

if((left1 == 1)&&(left2 == 1)&&(right1 == 0)&&(right2 == 0)) //1 1 0 0
      flag = 2;

if((left1 == 1)&&(left2 == 1)&&(right1 == 0)&&(right2 == 1)) //1 1 0 1
      flag = 0;

if((left1 == 1)&&(left2 == 1)&&(right1 == 1)&&(right2 == 0)) //1 1 1 0
      flag = 2;

if((left1 == 1)&&(left2 == 1)&&(right1 == 1)&&(right2 == 1)) //1 1 1 1
      flag = 3;
switch(flag)
{
case(0):Forward();break;

case(1):TurnRight();break;

case(2):TurnLeft();break;

default:Stop();break;
}
}

作者: zafu 时间: 2019-7-20 15:10
很不错，请楼主分享一下全部代码，1539429523@qq.com

作者: ?小萌新? 时间: 2019-10-29 17:18
楼主可不可以给我发一份完整代码，急用1078786909@qq.com

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