在两点之间绘制正方形

Question

我正在做一些练习，我已经坚持了几个小时了（对Java来说很陌生）。无论如何，这是我应该做的：当我运行程序时，我将在屏幕中间有一个正方形，当我单击该屏幕中的某个地方时，将在我单击的位置绘制另一个正方形，并且在这两点之间应该是 10 个正方形。因此，无论我单击何处，都应该始终绘制10个正方形。

但是，我无法使其正常运行。

这是我到目前为止设法做到的：

import se.lth.cs.ptdc.window.SimpleWindow;  
import se.lth.cs.ptdc.square.Square;

public class PrintSquares2 {

public static void main(String[] args) {
    SimpleWindow w = new SimpleWindow(600, 600, "PrintSquares2");
    int posX = 300;
    int posY = 300;
    int loop = 0;
    System.out.println("Skriv rotation");
    Square sq1 = new Square(posX,posY,200);
    sq1.draw(w);

            w.waitForMouseClick();
            int destX = w.getMouseX();
            int destY = w.getMouseY();
            System.out.println("Dest X: " + destX + " Dest Y: " + destY);
            System.out.println("Pos X: " + posX + " Pos Y: " + posY);
            SimpleWindow.delay(10);
            //sq1.erase(w);
            int jumpX = (destX - posX) / 10;
            int jumpY = (destY - posY) / 10;
            System.out.println(jumpX);

                while (posX < destX)
                {       
                    posX = posX+10;
                    SimpleWindow.delay(100);
                    loop++;
                    System.out.println("Loop: " + loop);
                    System.out.println("Dest X: " + destX + " Dest Y: " + destY);
                    System.out.println("Pos X: " + posX + " Pos Y: " + posY);       
                    Square sq2 = new Square(posX,posY,200);         
                    sq2.draw(w);                        
                }
                while (posX > destX)
                {
                    posX = posX-10;
                    SimpleWindow.delay(100);
                    loop++;
                    System.out.println("Loop: " + loop);
                    System.out.println("Dest X: " + destX + " Dest Y: " + destY);
                    System.out.println("Pos X: " + posX + " Pos Y: " + posY);
                    sq1.draw(w);
                    Square sq2 = new Square(posX,posY,200);         
                    sq2.draw(w);
                }
                while (posY < destY)
                {       
                    posY = posY+10;
                    SimpleWindow.delay(100);
                    loop++;
                    System.out.println("Loop: " + loop);
                    System.out.println("Dest X: " + destX + " Dest Y: " + destY);
                    System.out.println("Pos X: " + posX + " Pos Y: " + posY);
                    sq1.draw(w);
                    Square sq2 = new Square(posX,posY,200);         
                    sq2.draw(w);
                }
                while (posY > destY)
                {
                    posY = posY-10;
                    SimpleWindow.delay(100);
                    loop++;
                    System.out.println("Loop: " + loop);
                    System.out.println("Dest X: " + destX + " Dest Y: " + destY);
                    System.out.println("Pos X: " + posX + " Pos Y: " + posY);
                    sq1.draw(w);
                    Square sq2 = new Square(posX,posY,200);         
                    sq2.draw(w);
                }

            SimpleWindow.delay(10);
            sq1.draw(w);
            //SimpleWindow.clear(w);

    }

}

我很确定我把一切都搞得太复杂了，因为这应该是非常基本的。

最终结果应该看起来像这样：最终结果

Answer 1

这是我解决它的方法：

我不太理解se.lth.cs.ptdc.square.Square的文档，但我假设它在给定其左上角的坐标和边长的情况下绘制一个正方形。

因此，您拥有第一个正方形左上角的坐标和最后一个正方形中心的坐标。有了这个，不难得到最后一个正方形左上角的坐标：

lastX = centerX - side/2 lastY = centerY - side/2

有了它，你会发现起点和终点之间的差异：

diffX = posX - lastX diffY = posY - lastY

之后只需再画 9 个正方形：

for (int i=1; i<10; i++){
    squareX = posX + (diffX/10)*i;
    squareY = posY + (diffY/10)*i;
    Square square = new Square(squareX,squareY,200);         
    square.draw(w);
}

实际上，您做对了第一部分，只是搞砸了那些不必要的检查。希望对您有所帮助。

--
问候，SVZ。

Answer 2

同时更新 X 和 Y：

    int jumpX = (destX - posX) / 10;
    int jumpY = (destY - posY) / 10;
    if (posX > destX) {
        int temp = destX;
        destX = posX;
        posX = temp;
    }
    while (posX <= destX)
    {       
            SimpleWindow.delay(100);
            loop++;
            System.out.println("Loop: " + loop);
            System.out.println("Dest X: " + destX + " Dest Y: " + destY);
            System.out.println("Pos X: " + posX + " Pos Y: " + posY);       
            Square sq2 = new Square(posX,posY,200);         
            sq2.draw(w);                        
            posX = posX+jumpX;
            posY = posY+jumpY;
    }    
    SimpleWindow.delay(10);
    sq1.draw(w);

Answer 3

以下是您如何同时向两个方向移动（在对角线上）。

static final int Steps = 10;
private void test() {
  int x1 = 100;
  int y1 = 100;
  int x2 = 300;
  int y2 = 500;
  double dx = (double)(x2 - x1) / (double) Steps;
  double dy = (double)(y2 - y1) / (double) Steps;
  double x = x1;
  double y = x2;
  for ( int i = 0; i < Steps; i++) {
    // Simulate the drawing of the square.
    System.out.println("("+x+","+y+")");
    x += dx;
    y += dy;
  }
}