C#图形闪烁

Question

我正在研究一种绘图程序,但是在绘制橡皮带线时移动鼠标光标时出现闪烁问题.我希望你能帮助我删除那条闪烁的行,这里是代码:

using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Linq;
using System.Text;
using System.Windows.Forms;

namespace GraphicsTest
{
    public partial class Form1 : Form
    {
        int xFirst, yFirst;
        Bitmap bm = new Bitmap(1000, 1000);
        Graphics bmG;
        Graphics xG;
        Pen pen = new Pen(Color.Black, 1);
        bool draw = false;

        public Form1()
        {
            InitializeComponent();
        }

        private void Form1_Load(object sender, EventArgs e)
        {
            bmG = Graphics.FromImage(bm);
            xG = this.CreateGraphics();
            bmG.Clear(Color.White);
        }

        private void Form1_MouseDown(object sender, MouseEventArgs e)
        {
            xFirst = e.X;
            yFirst = e.Y;
            draw = true;
        }

        private void Form1_MouseUp(object sender, MouseEventArgs e)
        {
            bmG.DrawLine(pen, xFirst, yFirst, e.X, e.Y);
            draw = false;
            xG.DrawImage(bm, 0, 0);
        }

        private void Form1_MouseMove(object sender, MouseEventArgs e)
        {
            if (draw)
            {
                xG.DrawImage(bm, 0, 0);
                xG.DrawLine(pen, xFirst, yFirst, e.X, e.Y);
            }
        }

        private void Form1_Paint(object sender, PaintEventArgs e)
        {
            xG.DrawImage(bm, 0, 0);
        }
    }
}

Answer 1

首先不要使用,CreateGraphics()除非你绝对必须.将事件处理程序绑定到OnPaint并Invalidate()在要刷新曲面时调用.

如果你不想让它闪烁,你需要双重缓冲你的绘图表面.最简单的方法是将表单的DoubleBuffered属性设置为True.

如果你计划扩展它来绘制PictureBox控件,我强烈建议你.PictureBox默认为双缓冲,允许您更简单地控制绘图区域.

在代码中:

public partial class Form1 : Form
    {
    int xFirst, yFirst;
    Bitmap bm = new Bitmap(1000, 1000);
    Graphics bmG;
    Pen pen = new Pen(Color.Black, 1);
    bool draw = false;

    public Form1()
    {
        InitializeComponent();
    }

    private void Form1_Load(object sender, EventArgs e)
    {
        bmG = Graphics.FromImage(bm);
        bmG.Clear(Color.White);
    }

    private void Form1_MouseDown(object sender, MouseEventArgs e)
    {
        xFirst = e.X;
        yFirst = e.Y;
        draw = true;
    }

    private void Form1_MouseUp(object sender, MouseEventArgs e)
    {
        bmG.DrawLine(pen, xFirst, yFirst, e.X, e.Y);
        draw = false;
        Invalidate();
    }

    private void Form1_MouseMove(object sender, MouseEventArgs e)
    {
        if (draw)
        {
            Invalidate();
        }
    }

    private void Form1_Paint(object sender, PaintEventArgs e)
    {
        if (draw) {
            e.Graphics.DrawImage(bm, 0, 0);
            e.Graphics.DrawLine(pen, xFirst, yFirst, e.X, e.Y);
        } else {
            e.Graphics.DrawImage(bm, 0, 0);
        }
    }
}

编辑:

另一个问题是,您正在创建一个私有Pen成员.笔(和Brushes,以及许多GDI +对象)表示需要处理的非托管对象的句柄,否则您的程序将泄漏.将它们包装在using语句中(首选和异常安全的方式)或者在表单的Dispose方法中明确地处理它们.

或者在System.Drawing中,您可以访问一些不需要(也不应该)处置的预构建的笔和画笔.使用它们像:

    private void Form1_Paint(object sender, PaintEventArgs e)
    {
        if (draw) {
            e.Graphics.DrawImage(bm, 0, 0);
            e.Graphics.DrawLine(Pens.Black, xFirst, yFirst, e.X, e.Y);
        } else {
            e.Graphics.DrawImage(bm, 0, 0);
        }
    }

Answer 2

它闪烁的原因是你正在绘制背景(立即显示在屏幕上,擦掉线条),然后叠加线条.因此,线条不断消失和出现,呈现闪烁的显示.

对此的最佳解决方案称为双缓冲.你所做的是将整个图像绘制到"屏幕外"位图,并在完成时仅在屏幕上显示.因为您只显示完成的图像,所以没有闪烁效果.您应该能够设置this.DoubleBuffered = true以使WinForms为您完成所有艰苦的工作.

注意:你不应该真的在绘图处理程序之外绘图 - 理想情况下,你应该Invalidate()需要重绘的区域,然后你的绘图处理程序将重绘该区域(根据需要重叠任何线条等).