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我正在构建一个 3d（红色，青色）程序。进展顺利，但在组合时，“图像”（见代码）似乎没有受到影响。

我的错误是什么？

认为： 

RImage_P(point) = {5,5}
startCyan(point) = {0,0}
startRed(point) = {5,5}

CImage，RImage位于代码下方。图像结果是全黑透明图像。对于测试和调试，我注释了Alpha（结果是一样的）。

private void bitCombine(System.Drawing.Point startCyan, System.Drawing.Point startRed)
{
    using (Bitmap image = new Bitmap(CImage.Width + Math.Abs(RImage_P.X), CImage.Height + Math.Abs(RImage_P.Y),PixelFormat.Format32bppArgb))
    {
        var rectfull = new Rectangle(0, 0, image.Width, image.Height);
        var rect = new Rectangle(0, 0, CImage.Width, CImage.Height);
        var bitsC = CImage.LockBits(rect, ImageLockMode.ReadOnly
                                    , PixelFormat.Format32bppArgb);
        var bitsA = alpha.LockBits(rect, ImageLockMode.ReadOnly
                                    , PixelFormat.Format32bppArgb);
        var bitsOutput = image.LockBits(rectfull, ImageLockMode.WriteOnly
                                    , PixelFormat.Format32bppArgb);
        unsafe
        {
            for (int y = 0; y < CImage.Height; y++)
            {
                byte* ptrC = (byte*)bitsC.Scan0 +  y * bitsC.Stride;
                byte* ptrA = (byte*)bitsA.Scan0 +  y * bitsA.Stride;
                byte* ptrOutput = (byte*)bitsOutput.Scan0 
                                    + (y+startCyan.Y)* bitsOutput.Stride;
                for (int x = 0; x < CImage.Width; x++)// cyan values
                {// gui+ is reversed BGRA
                    ptrOutput[4 * (x+startCyan.X)] = ptrC[4 * x];           // blue
                    ptrOutput[4 * (x + startCyan.X) + 1] = ptrC[4 * x + 1];// green
                    //ptrOutput[4 * (x + startCyan.X) + 3] = ptrA[4 * x + 3];// alpha
                }
            }
            CImage.UnlockBits(bitsC);

            var bitsR = RImage.LockBits(rect, ImageLockMode.ReadOnly
                                        , PixelFormat.Format32bppArgb);
            for (int y = 0; y < RImage.Height; y++)// red values
            {
                byte* ptrR = (byte*)bitsR.Scan0 + y  * bitsR.Stride;
                byte* ptrA = (byte*)bitsA.Scan0 + y  * bitsA.Stride;
                byte* ptrOutput = (byte*)bitsOutput.Scan0
                                    + (y+ startRed.Y) * bitsOutput.Stride;
                for (int x = 0; x < RImage.Width; x++)
                {// gui+ is reversed BGRA
                    ptrOutput[4 * (x + startRed.X) + 2] = ptrR[4 * x + 2];   // red
                    //ptrOutput[4 * (x + startRed.X) + 3] = ptrA[4 * x + 3];// alpha
                }
            }

            RImage.UnlockBits(bitsR);
            alpha.UnlockBits(bitsA);
            image.UnlockBits(bitsOutput);


            this.image = image;
        }
    }
}

图像：

影音 图像

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1 回答 1

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我用图像可视化器发现它确实有效，但 alpha 通道 =0 答案是将 alpha 通道的值更改为 255。

于 2012-09-01T14:40:37.820 回答