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標題: 基于51單片機的智能紅外循跡小車程序 [打印本頁]

作者: cqupt.qwert 時間: 2019-6-14 19:17
標題: 基于51單片機的智能紅外循跡小車程序
最近剛結(jié)束單片機講座的課，在寫關(guān)于智能汽車的文章的同時自己做了一個簡單的智能循跡小車，這里采用的是光電循跡模塊，使用比較簡單，
紅外對管輸出的電壓模擬信號通過比較器轉(zhuǎn)換為了數(shù)字電平，只需要根據(jù)輸出的電平高低控制電機的正反轉(zhuǎn)，這邊附上部分單片機代碼

#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 時間: 2019-7-20 15:10
很不錯，請樓主分享一下全部代碼，1539429523@qq.com

作者: ?小萌新? 時間: 2019-10-29 17:18
樓主可不可以給我發(fā)一份完整代碼，急用1078786909@qq.com

歡迎光臨 (http://m.raoushi.com/bbs/)