我正在尝试用Java实现DDA算法来绘制一条线。 Line Rasterizer实现了一个接口。然后我在名为rasterizeLine
的类中调用Canva
函数。 LineRasterizes成功获得x和y点。没有错误被抛出,但是算法必须存在一些逻辑问题,因为没有绘制线。你能帮我找到合乎逻辑的错误吗?
接口
package rasterops;
import rasterdata.RasterImage;
public interface LineRasterizer<PixelType> {
RasterImage<PixelType> rasterizeLine(RasterImage<PixelType> img,
double x1, double y1, double x2, double y2,
PixelType value);
}
Line Rasterizer
package rasterops;
import rasterdata.RasterImage;
public class LineRasterizerDDA<PixelType> implements LineRasterizer <PixelType> {
@Override
public RasterImage<PixelType> rasterizeLine(RasterImage<PixelType> img, double x1, double y1, double x2, double y2, PixelType value) {
double dy = y2-y1;
double dx = x2-x1;
double k = dy/dx;
double y = y1;
RasterImage<PixelType> result = img;
for( double x = x1; x <= x2; x++ ){
result = result.withPixel((int)x, (int)Math.round(y), value);
y = y1 + k;
}
return result;
}
}
Canva - 调用函数
panel.addMouseListener(new MouseAdapter() {
@Override
public void mousePressed(MouseEvent e) {
previousX = e.getX();
previousY = e.getY();
}
});
panel.addMouseMotionListener(new MouseAdapter() {
@Override
public void mouseDragged(MouseEvent e) {
final double startX = previousX / (panel.getWidth() - 1.0); //k zamysleni: proc 1.0 a ne 1?
final double startY = 1 - previousY / (panel.getHeight() - 1.0);
final double endX = e.getX() / (panel.getWidth() - 1.0);
final double endY = 1 - e.getY() / (panel.getHeight() - 1.0);
clear(); // zkuste zakomentovat
rasterImage = liner.rasterizeLine(rasterImage,
startX, startY, endX, endY,
0xffff00);
panel.repaint();
}
});
答案 0 :(得分:3)
这一行:
y = y1 + k;
继续为y
分配相同的值,因为y1
和k
并未发生变化。
Example output, drawing line between (0,0) and (10, 10):
O..........
.OOOOOOOOOO
...........
...........
...........
...........
...........
...........
...........
...........
...........
你的意思是:
y = y + k;
或y += k;
。
Example output, drawing line between (0,0) and (10, 10):
O..........
.O.........
..O........
...O.......
....O......
.....O.....
......O....
.......O...
........O..
.........O.
..........O