我正在为一个高中项目设计和编程一个类似电梯的机器人。我能做点什么让这更简单吗？或更好？我附上了我在 AutoCAD Inventor 中制作的带有标签的设计图片。

对于那些不熟悉 RobotC 或 VEX(它与 C 和 C++ 非常相似)的人:限位开关(limit1、limit2、...)和碰撞开关(floor1、floor2、...)是模拟按钮和如果未按下则返回值 0，如果按下则返回 1。电机(主电机)旋转齿轮，使机构在 slider 上向上移动。当伸出电机机构的轴上下移动时，它会按下限位开关并使其返回值 1。

int callup [3];
int calldown [3];
int floorat[3];

    int main ()
    {

        if (SensorValue[limit1] == 1)
        {
            floorat[0] = 1;
        }
        else
        {
            floorat[0] = 0;
        }

        if (SensorValue[limit2] == 1)
        {
            floorat[1] = 1;
        }
        else
        {
            floorat[1] = 0;
        }

        if (SensorValue[limit3] == 1)
        {
            floorat[2] = 1;
        }
        else
        {
            floorat[2] = 0;
        }

        if (SensorValue[floor1] == 1)
        {
            calldown[0] = 1;
            SensorValue[LED1] = 1;
        }

        if (SensorValue[floor2] == 1 && floorat[2] == 1)
        {
            calldown[1] = 1;
            SensorValue[LED2] = 1;
        }

        if (SensorValue[floor2] == 1 && floorat[0] == 1)
        {
            callup[1] = 1;
            SensorValue[LED2] = 1;
        }

        if (SensorValue[floor3])
        {
            callup[2] = 1;
            SensorValue[LED3] = 1;
        }

        motors ();

    }


    void motors ()
    {

        if (callup[2] == 1 && floorat[2] == 1)
        {
            int x = 1;
            while (x < 3)
            {
                SensorValue[LED3] = 1;
                wait(0.5);
                SensorValue[LED3] = 0;
                wait(0.5);
            }
            callup[2] = 0;
            main ();
        }
        else if (callup[1] == 1 && floorat[1] == 1)
        {
            int x = 1;
            while (x < 3)
            {
                SensorValue[LED2] = 1;
                wait(0.5);
                SensorValue[LED2] = 0;
                wait(0.5);
            }
            callup[1] = 0;
            main ();
        }
        else if (callup[0] == 1 && floorat[0] == 1)
        {
            int x = 1;
            while (x < 3)
            {
                SensorValue[LED1] = 1;
                wait(0.5);
                SensorValue[LED1] = 0;
                wait(0.5);
            }
            callup[0] = 0;
            main ();
        }

        if (callup[2] == 1 && floorat[1] == 1 && calldown[0] == 0 || callup[2] == 1 && floorat[0] == 1 && callup[1] == 0)
        {
            startMotor(mainMotor, 60);
            untilTouch(limit3);
            stopMotor(mainMotor);
            callup[2] = 0;
            wait(1);
            main ();
        }

        if (callup[1] == 1 && floorat[0] == 1)
        {
            startMotor(mainMotor, 60);
            untilTouch(limit2);
            stopMotor(mainMotor);
            callup[1] = 0;
            wait(1);
            main();
        }

        if (calldown[1] == 1 && floorat[2] == 1)
        {
            startMotor(mainMotor, -60);
            untilTouch(limit2);
            stopMotor(mainMotor);
            calldown[1] = 0;
            wait(1);
            main();
        }

        if (calldown[0] == 1 && floorat[2] == 1 && calldown[1] == 0 || calldown[0] == 1 && floorat[1] == 1)
        {
            startMotor(mainMotor, -60);
            untilTouch(limit1);
            stopMotor(mainMotor);
            calldown[0] = 0;
            wait(1);
            main();
        }
    }

虽然这个问题应该不用关心，但是startMotor命令中的60是电机的转速，只是为了说的清楚一点。

如有任何问题，请随时提出。

最佳答案

让我们定义电梯在给定时刻的状态:

电梯可以上行、下行或空闲。

电梯在给定的楼层，并在触发开关时从一层转到另一层:

现在，如果我们将其转换为一些伪代码(应该很容易转换为 RobotC):

enum elevator_status = { idle, down, up };
int currentfloor; //1, 2, 3 


switch(elevator_status)
{
    case idle:    
        //we check if a button is pressed and possibly go up or down           
        if(SensorValue(floor1))
        {         
             if(currentfloor > 1)
                elevator_status = down;               
        }
        else if(SensorValue(floor2))
        {
              if(currentfloor > 2)
                  elevator_status = down;
              else if(currentfloor < 2)
                  elevator_status = up;               
        }
        else if(SensorValue(floor3))
        {
              if(currentfloor < 3)
                  elevator_status = up;   
        }
        break;

    case up:
    case down:    
        //we check if we trigger a floor switch and stop the elevator
        if(SensorValue(limit1))
        {      
           currentfloor = 1;
           elevator_status = idle;
        }
        else if(SensorValue(limit2))
        {
           currentfloor = 2;
           elevator_status = idle;
        }
        else if(SensorValue(limit3))
        {
           currentfloor = 3;
           elevator_status = idle;
        }
        break;
}


//we set the speed of the motor
if(elevator_status == up)
{
   set_motorstate(cw);
)
else if(elevator_status == down)
{
   set_motorstate(ccw);   
}
else if(elevator_status == idle)
{
   set_motorstate(idle);   
}

注意:在这段代码中，电梯只会在电梯空闲时处理新的上下楼层调用。它在移动时不存储上行和下行调用，稍后再去那里。我不知道这是否是您的要求。

关于c++ - RobotC - 电梯编程，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/19883220/