DDA算法画直线有个问题，麻烦大家看看，帮帮忙？

[timmyangel]

private void DDAToolStripMenuItem_Click(object sender, EventArgs e)
{

if (bt == null)
bt = new Bitmap(PictureBox1.Width,PictureBox1.Height);

float xa = 33;
float ya = 44;
float xb = 66;
float yb = 234;

float delta_x, delta_y;
float x, y;
int dx, dy, steps, k;
dx = (int)(xb - xa);
dy = (int)(yb - ya);
if (Math.Abs(dx) > Math.Abs(dy))
steps = Math.Abs(dx);
else steps = Math.Abs(dy);
delta_x = (float)dx / (float)steps;
delta_y = (float)dy / (float)steps;
x = xa;
y = ya;
bt.SetPixel((int)x, (int)y, BtnColor.BackColor);
for (k = 1; k <= steps; k++)
{
x += delta_x;
y += delta_y;
bt.SetPixel((int)x, (int)y, BtnColor.BackColor);
}
PictureBox1.Image = bt;


}
这是之前的DDA画线，但只能定位坐标来画，我问下怎么样才能用鼠标在PictureBox1上面随便画直线？？
已知：MouseDown { pStart.X = e.X;
pStart.Y = e.Y;}
MouseUp{ pEnd.X = e.X;
pEnd.Y = e.Y;}

[h_w_king]

if (bt == null)
bt = new Bitmap(pictureBox1.Width, pictureBox1.Height);

Graphics g = Graphics.FromImage(bt);
g.DrawLine(Pens.Black, new Point(pStart.X, pStart.Y ), new Point(pEnd.X, pEnd.Y));

this.pictureBox1.Image = bt;

[LorenLiu]

[Quote=引用 1 楼 h_w_king 的回复:]
if (bt == null)
bt = new Bitmap(pictureBox1.Width, pictureBox1.Height);

Graphics g = Graphics.FromImage(bt);
g.DrawLine(Pens.Black, new Point(pStart.X, pStart.Y ), new Point(pEnd.X, pEnd.Y));

this.pictureBox1.Image = bt;
[/Quote]

LZ是要用DDA方法来画线。。。。

LZ可以把DDA方法定义成一个函数 Bitmap DDA（Point startPoint, Point endPoint)来实现，startPoint和endPoint分别传入鼠标的两个位置就好了

[wartim]

我以前写的，在form上随便画线
第一次点击确定起点
第二次点击确定终点，而且在鼠标移动过程中也会动态的显示直线
第三次点击画的直线消失

namespace WindowsApplication27
{
    public partial class Form1 : Form
    {
        List<Point> Points = new List<Point>();
        Point MovePoint;
        public Form1()
        {
            InitializeComponent();
        }

        private void Form1_MouseDown(object sender, MouseEventArgs e)
        {
            if (Points.Count == 0)
            {
                Points.Add(e.Location);
                MovePoint = new Point(e.X, e.Y);
            }
            else if (Points.Count == 1)
                Points.Add(e.Location);
            else if (Points.Count == 2)
            {
                Points.RemoveAt(0);
                Points.RemoveAt(0);
            }
        }

        private void Form1_MouseMove(object sender, MouseEventArgs e)
        {
            if (Points.Count == 1)
                MovePoint = new Point(e.X,e.Y);
            this.Refresh(); 
        }

        private void Form1_Paint(object sender, PaintEventArgs e)
        {
            if (Points.Count == 1)
                e.Graphics.DrawLine(new Pen(new SolidBrush(Color.Red)), Points[0], MovePoint);
            else if (Points.Count == 2)
                e.Graphics.DrawLine(new Pen(new SolidBrush(Color.Red)), Points[0], Points[1]);
        }
    }
}

[timmyangel]

[Quote=引用 2 楼 LorenLiu 的回复:]
引用 1 楼 h_w_king 的回复:
if (bt == null)
bt = new Bitmap(pictureBox1.Width, pictureBox1.Height);

Graphics g = Graphics.FromImage(bt);
g.DrawLine(Pens.Black, new Point(pStart.X, pStart.Y ), new Point(pEnd.X, pEnd.Y));

this.pictureBox1.Image = bt;


LZ是要用DDA方法来画线。。。。

LZ可以把DDA方法定义成一个函数 Bitmap DDA（Po…
[/Quote]
怎么样把鼠标的两个位置传入startPoint和endPoint 呢？之前弄过还是不行？

[LorenLiu]

[Quote=引用 4 楼 timmyangel 的回复:]
引用 2 楼 LorenLiu 的回复:
引用 1 楼 h_w_king 的回复:
if (bt == null)
bt = new Bitmap(pictureBox1.Width, pictureBox1.Height);

Graphics g = Graphics.FromImage(bt);
g.DrawLine(Pens.Black, new Point(pStart.X, pStart.Y ), new Point(pEnd.X, pEnd.Y));

this.pictureBox1.Image = bt;


LZ是要用DDA方法来画线。。。。

LZ可以把DDA方法定…
[/Quote]

PictureBox1注册MouseDown和MouseUp事件后，就能得到e.Location，就是需要的坐标值啊

[timmyangel]

假如我现在定义了这样的一个类，怎么样把值传进去呢？
public class Pixel
{
public int _x;
public int _y;

public Pixel()
{
this._x = 0;
this._y = 0;
}

public Pixel(int x,int y)
{
this._x = x;
this._y = y;
}
}
public class Point
{
public float x;
public float y;
}
public Pixel LineDDA(Point p0,Point p1,Pixel precedent)
{
Pixel descendent = new Pixel();
float k=Math.Abs((p1.y-p0.y)/(p1.x-p0.x));

if(k>1||k==1)
{
descendent._x = precedent._x + 1;
descendent._y = precedent._y + 1;
}
else
{
descendent._x = precedent._x + 1;
descendent._y = precedent._y + 1;
}
return descendent;
}
MouseDown { pStart.X = e.X;
pStart.Y = e.Y;}
MouseUp{ pEnd.X = e.X;
pEnd.Y = e.Y;}
之前试过老是不行，大家帮下忙......课程设计来的

[LorenLiu]

已经弄出来了，怎么发给你？

[timmyangel]

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