問題描述

大家好.在厭倦了 Get/Set Pixel 和 RotateTransfom 的緩慢性能和古怪行為之后，我求助于使用 LockBits 進行 2d 位圖圖像旋轉.所以這是我想出的代碼，據我估計，它應該可以完美運行.沒有.

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);
        }

有人有什么想法嗎?我剛出來.提前致謝！

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

這就是我最終使用的(非常感謝 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;
    }

這和他的一樣，只是基于每個圖像，而不是形式.

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

推薦答案

您正在為自己挖一個更深的坑.這很早就出錯了，旋轉位圖的大小不是寬 x 高.這也是非常低效的.您需要啟動 RotateTransform，同時使用 TranslateTransform 并選擇正確的圖像繪制位置也很重要.

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.

這是一個示例 Windows 窗體應用程序，它圍繞其中心點旋轉位圖，偏移量剛好足以在窗體旋轉時接觸窗體的內邊緣.在表單上放置一個計時器并使用項目 + 屬性、資源選項卡添加圖像資源.將其命名為 SampleImage，它不必是方形的.使代碼看起來像這樣:

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);
    }
}

您可以通過創建 32bppPArgb 格式的位圖來加快繪制速度，我跳過了這一步.

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

這篇關于LockBits 圖像旋轉方法不起作用?的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！