㈠ c语言设计机器人运动轨迹程序……期末急需!!!
几个问题:
1.机器人 电机转速有反馈吗?应该可以知道什么时候碰倒障碍物的吧?
2.是不是必须通过黄线找到红色区域啊?能不能直接找红色区域?
3.机器人肯定有两个motor的把,左右实现转向对吧?
运动轨迹 应该是找到黄线后,机器人的动作只有两种 带速左转右转,当前位置在黄线上,那么右转,转到黑色区域左转;同理左转转到黑色区域就右转,最后实现动作像蛇一样沿着黄线前进。速度是根据电机前进方向的速度决定的,但是如果光感采样频率太低,电机速度太快有可能甩出黄线区域。
这个逻辑用C应该很容易实现,难点是怎么快速找到红色区域。
㈡ 机器人 C语言
这样写吧,清楚一点。
我觉得我这样定义的话会让程序清楚很多。
typedef struct
{
int x;
int y;
} pos_stru;
void main(void)
{
int cmd;
int face[4]={0,1,2,3}; //x+,y+,x-,y- 定义行进方向
int face_start=face[0]; //初始方向
pos_stru start={0,0}; //开始位置
while(scanf("%d",&cmd)!=NULL)
{
if(cmd==0)
break;
if(cmd<-2) //错误处理
{
printf("Invide input.");
}
switch(cmd) //先处理方向问题-1/-2
{
case -1: //左转
if(face_start<3)
face_start=face[face_start+1];
else
face_start=face[0];
break;
case -2: //右转
if(face_start>0)
face_start=face[face_start-1];
else
face_start=face[3];
break;
default:
break;
}
if(cmd>0) //不是方向的处理
switch(face_start)
{
case 0:
start.x+=cmd;
break;
case 1:
start.y+=cmd;
break;
case 2:
start.x-=cmd;
break;
case 3:
start.y-=cmd;
break;
default:
break;
}
//printf("tmp out:x=%d,y=%d\n",start.x,start.y);
}
printf("result:x=%d,y=%d\n",start.x,start.y);
}
out put:
-1
2
-2
1
0
result:x=1,y=2
Press any key to continue
问题补充:
没给定义应该是都初始化为0了吧~
答:不对。随机的这也是我们要求C语言必须初值的原因。下面是试验
void main(void)
{
int a;
int b=0;
printf("a=%d,b=%d\n",a,b);
}
output:
a=-858993460,b=0
Press any key to continue
㈢ c语言问题,一个机器人只在横坐标上移动,要求输入开始坐标,结束坐标,移动的步骤,输出机器人最终停止
#include <stdio.h>
#include <conio.h>
void main( )
{
int StartPoint, EndPoint, move[1000], CurrentPoint, i;
printf("Enter starting address: ");
scanf("%d", &StartPoint);
printf("Enter ending address: ");
scanf("%d", &EndPoint);
while(EndPoint < StartPoint)
{
printf(" Error! ending address must be >= %d ", StartPoint);
printf("Enter starting address: ");
scanf("%d", &EndPoint);
}
printf("Enter values to move now: ");
i = 0;
scanf("%d", &move[i]);
while(move[i] != 0) scanf("%d", &move[++i]);
for(CurrentPoint = StartPoint, i = 0; move[i]; i++)
{
CurrentPoint += move[i];
if(CurrentPoint > EndPoint) CurrentPoint = EndPoint;
if(CurrentPoint < StartPoint) CurrentPoint = StartPoint;
}
printf(" Final location: %d", CurrentPoint);
getch( ):
}
运行结果
㈣ c语言问题,一个机器人只在横坐标上移动,要求输入开始坐标,结束坐标,移动的步骤,输出机器人最终停止
看到楼主的修改后的代码,我给出一些修改意见
#include <stdio.h>
/*#include <math.h>*/ /*这行你是将conio.h改为math.h,其实这两个都可删去,只要不用getch*/
int main( ) /*将返回类型由void改为int,可以*/
{
int StartPoint, EndPoint, CurrentPoint, move; /*为了可读性,将你的i改为move了*/
printf("Enter starting address: ");
scanf("%d", &StartPoint);
printf("Enter ending address: ");
scanf("%d", &EndPoint);
while(EndPoint < StartPoint)
{
printf("\nError! ending address must be >= %d\n\n", StartPoint);
printf("Enter ending address: "); /*你的这句改得好*/
scanf("%d", &EndPoint);
}
printf("Enter values to move now: ");
/* i = 0; //这里的连续9行有问题,将其修改后见下面的注。
scanf("%d", &i);
while(i != 0) scanf("%d", &i);
for(CurrentPoint = StartPoint, i = 0; move[i]; i++)
{
CurrentPoint += i;
if(CurrentPoint > EndPoint) CurrentPoint = EndPoint;
if(CurrentPoint < StartPoint) CurrentPoint = StartPoint;
} */
printf("\nFinal location: %d", CurrentPoint);
return (0); /*如果返回类型是int,则需要这行,如果是void,则不需这行*/
}
注:
CurrentPoint = StartPoint;
scanf("%d", &move);
while(move != 0)
{
CurrentPoint += move;
if(CurrentPoint > EndPoint) CurrentPoint = EndPoint;
if(CurrentPoint < StartPoint) CurrentPoint = StartPoint;
scanf("%d", &i);
}
㈤ C语言控制机器人走路的程序
//设置左右马达参数
void ZYMotor(unsigned char ZState, unsigned char ZSpeed, unsigned char YState, unsigned char YSpeed)
{
SetMotor(_MOTOR_ZuoMotor_, ZState, ZSpeed);
SetMotor(_MOTOR_YouMotor_, YState, YSpeed);
}
//确认各个端口的状态
void ShiBieXian()
{
n=1;
XJ[1]= GetADScable(_SCABLEAD_HB1_);
XJ[2]= GetADScable(_SCABLEAD_HB2_);
XJ[3]= GetADScable(_SCABLEAD_HB3_);
XJ[4]= GetADScable(_SCABLEAD_HB4_);
XJ[5]= GetADScable(_SCABLEAD_HB5_);
while(n<=5)
{
if(XJ[n]>FZ[n])
{HB[n]=0;}
else
{HB[n]=1;}
n++;
}
}
void GoOnLine()
{
if(HB[2]&&HB[3]&&HB[4])
{ZYMotor(0,60,0,60);}
else
{
if(HB[2]&&HB[3])
{ZYMotor(2,20,0,80);}
else
{
if (HB[2])
{ZYMotor(2,50,0,80);}
else
{
if(HB[3]&&HB[4] )
{ZYMotor(0,80,2,20);}
else
{
if(HB[4])
{ZYMotor(0,80,2,50);}
else
{
if (HB[1]&&!HB[2]&&!HB[3]&&!HB[4]&&!HB[5])
{ZYMotor(2, 80, 0, 80);}
else
{
if (!HB[1]&&!HB[2]&&!HB[3]&&!HB[4]&&HB[5])
{ZYMotor(0, 80, 2, 80);}
else
{ZYMotor(0, 60, 0, 60);}
}
}
}
}
}
}
}