LockBits 图像旋转方法不起作用?

时间:2022-11-23
本文介绍了LockBits 图像旋转方法不起作用?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

大家好.在厌倦了 Get/Set Pixel 和 RotateTransfom 的缓慢性能和古怪行为之后,我求助于使用 LockBits 进行 2d 位图图像旋转.所以这是我想出的代码,据我估计,它应该可以完美运行.没有.

Hey all. I resorted to using LockBits for 2d bitmap image rotation after getting fed up with the slow performance and wacky behavior of both Get/Set Pixel, and RotateTransfom. So here is the code I've come up with, and by my reckoning, it should work perfectly. It doesn't.

private static void InternalRotateImage(Bitmap originalBitmap, Bitmap rotatedBitmap, PointF centerPoint, float theta)
    {
        BitmapData originalData = originalBitmap.LockBits(
            new Rectangle(0, 0, originalBitmap.Width, originalBitmap.Height),
            ImageLockMode.ReadOnly,
            originalBitmap.PixelFormat);

        BitmapData rotatedData = rotatedBitmap.LockBits(
            new Rectangle(0, 0, rotatedBitmap.Width, rotatedBitmap.Height),
            ImageLockMode.WriteOnly,
            rotatedBitmap.PixelFormat);

        unsafe
        {
            byte[,] A = new byte[originalData.Height * 2, originalBitmap.Width * 2];
            byte[,] R = new byte[originalData.Height * 2, originalBitmap.Width * 2];
            byte[,] G = new byte[originalData.Height * 2, originalBitmap.Width * 2];
            byte[,] B = new byte[originalData.Height * 2, originalBitmap.Width * 2];

            for (int y = 0; y < originalData.Height; y++)
            {
                byte* row = (byte*)originalData.Scan0 + (y * originalData.Stride);
                for (int x = 0; x < originalData.Width; x++)
                {
                    B[y, x] = row[x * 4];
                    G[y, x] = row[x * 4 + 1];
                    R[y, x] = row[x * 4 + 2];
                    A[y, x] = row[x * 4 + 3];
                }
            }

            for (int y = 0; y < rotatedData.Height; y++)
            {
                byte* row = (byte*)rotatedData.Scan0 + (y * rotatedData.Stride);
                for (int x = 0; x < rotatedData.Width; x++)
                {
                    int newy = (int)Math.Abs((Math.Cos(theta) * (x - centerPoint.X) - Math.Sin(theta) * (y - centerPoint.Y) + centerPoint.X));
                    int newx = (int)Math.Abs((Math.Sin(theta) * (x - centerPoint.X) + Math.Cos(theta) * (y - centerPoint.Y) + centerPoint.Y));

                    row[x * 4] = B[newy, newx];
                    row[x * 4 + 1] = G[newy, newx];
                    row[x * 4 + 2] = R[newy, newx];
                    row[x * 4 + 3] = A[newy, newx];
                }
            }

        }
            originalBitmap.UnlockBits(originalData);
            rotatedBitmap.UnlockBits(rotatedData);
        }

有人有什么想法吗?我刚出来.提前致谢!

Anybody got any ideas? I'm fresh out. Thanks in advance!

这就是我最终使用的(非常感谢 Hans Passant):

This is what I ended up using (many thanks to Hans Passant):

private Image RotateImage(Image img, float rotationAngle)
    {
        Image image = new Bitmap(img.Width * 2, img.Height * 2);
        Graphics gfx = Graphics.FromImage(image);

        int center = (int)Math.Sqrt(img.Width * img.Width + img.Height * img.Height) / 2;
        gfx.TranslateTransform(center, center);
        gfx.RotateTransform(rotationAngle);
        gfx.DrawImage(img, -img.Width / 2, -img.Height / 2);

        return image;
    }

这和他的一样,只是基于每个图像,而不是形式.

It's the same thing as his, just on a per image basis, as opposed to a form.

推荐答案

您正在为自己挖一个更深的坑.这很早就出错了,旋转位图的大小不是宽 x 高.这也是非常低效的.您需要启动 RotateTransform,同时使用 TranslateTransform 并选择正确的图像绘制位置也很重要.

You are digging yourself a deeper hole. This goes wrong early, the size of the rotated bitmap is not Width x Height. It is also very inefficient. You need to get RotateTransform going, it is important to also use TranslateTransform and pick the correct image drawing location.

这是一个示例 Windows 窗体应用程序,它围绕其中心点旋转位图,偏移量刚好足以在窗体旋转时接触窗体的内边缘.在表单上放置一个计时器并使用项目 + 属性、资源选项卡添加图像资源.将其命名为 SampleImage,它不必是方形的.使代码看起来像这样:

Here's a sample Windows Forms app that rotates a bitmap around its center point, offset just enough to touch the inner edge of the form when it rotates. Drop a Timer on the form and add an image resource with Project + Properties, Resource tab. Name it SampleImage, it doesn't have to be square. Make the code look like this:

public partial class Form1 : Form {
    private float mDegrees;
    private Image mBmp;
    public Form1() {
        InitializeComponent();
        mBmp = Properties.Resources.SampleImage;
        timer1.Enabled = true;
        timer1.Interval = 50;
        timer1.Tick += new System.EventHandler(this.timer1_Tick);
        this.DoubleBuffered = true;
    }
    private void timer1_Tick(object sender, EventArgs e) {
        mDegrees += 3.0F;
        this.Invalidate();
    }
    protected override void OnPaint(PaintEventArgs e) {
        int center = (int)Math.Sqrt(mBmp.Width * mBmp.Width + mBmp.Height * mBmp.Height) / 2;
        e.Graphics.TranslateTransform(center, center);
        e.Graphics.RotateTransform(mDegrees);
        e.Graphics.DrawImage(mBmp, -mBmp.Width/2, -mBmp.Height/2);
    }
}

您可以通过创建 32bppPArgb 格式的位图来加快绘制速度,我跳过了这一步.

You can make draw a lot faster by creating a bitmap in the 32bppPArgb format, I skipped that step.

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