Question

我有一个从相机捕获的RGB图像（RGB 4：4：4色彩空间，每像素24位）。我使用Gorgon 2D库（基于SharpDX构建）将该图像显示为纹理，因此我必须将其转换为ARGB。我使用此代码（不是我的代码）将RGB相机图像转换为RGBA。

[StructLayout(LayoutKind.Sequential)]
public struct RGBA
{
            public byte r;
            public byte g;
            public byte b;
            public byte a;
}

[StructLayout(LayoutKind.Sequential)]
public struct RGB
{
            public byte r;
            public byte g;
            public byte b;
}

unsafe void internalCvt(long pixelCount, byte* rgbP, byte* rgbaP)
            {
                for (long i = 0, offsetRgb = 0; i < pixelCount; i += 4, offsetRgb += 12)
                {
                    uint c1 = *(uint*)(rgbP + offsetRgb);
                    uint c2 = *(uint*)(rgbP + offsetRgb + 3);
                    uint c3 = *(uint*)(rgbP + offsetRgb + 6);
                    uint c4 = *(uint*)(rgbP + offsetRgb + 9);
                    ((uint*)rgbaP)[i] = c1 | 0xff000000;
                    ((uint*)rgbaP)[i + 1] = c2 | 0xff000000;
                    ((uint*)rgbaP)[i + 2] = c3 | 0xff000000;
                    ((uint*)rgbaP)[i + 3] = c4 | 0xff000000;
                }
            }

public unsafe void RGB2RGBA(int pixelCount, byte[] rgbData, byte[] rgbaData)
            {
                if ((pixelCount & 3) != 0) throw new ArgumentException();
                fixed (byte* rgbP = &rgbData[0], rgbaP = &rgbaData[0])
                {
                    internalCvt(pixelCount, rgbP, rgbaP);
                }
            }

然后像这样将RGB转换为RGBA：

byte[] rgb = new byte[800*600*3]; //Stored data
byte[] rgba = new byte[800 * 600 * 4];
RGB2RGBA(800*600, rgb, rgba)

然后我将rgba用作Gorgon纹理的数据：

unsafe
{
     fixed(void* rgbaPtr = rgba)
     {
          var buff = new GorgonNativeBuffer<byte>(rgbaPtr, 800*600*4);                      
          GorgonImageBuffer imb = new GorgonImageBuffer(buff, 800, 600, BufferFormat.R8G8B8A8_UNorm);

          //Set Texture data  GorgonTexture2D                                
          Texture.SetData(imb, new SharpDX.Rectangle(0, 0, 800, 600), 0, 0, CopyMode.NoOverwrite);
         }
}

但是纹理图像的颜色不像相机图像的颜色。

所以我想我必须将相机图像转换为ARGB（而不是RGBA），以便它可以显示Gorgon纹理，但是我不知道如何使用上面的代码来实现。你们能给我一些提示吗？

谢谢！

下面是我在上面的代码中使用的Gorgon库类型的链接

GorgonTexture2D GorgonNativeBuffer GorgonImageBuffer

Answer 1

目标的字节顺序在内存中为R G B A。源的字节顺序在内存中为B G R。因此，除了将每3个字节扩展到4个字节并将FF放入新的A通道外，R和B通道还需要交换位置。例如，

unsafe void internalCvt(long pixelCount, byte* rgbP, uint* rgbaP)
{
    for (long i = 0, offsetRgb = 0; i < pixelCount; i += 4, offsetRgb += 12)
    {
        uint c1 = *(uint*)(rgbP + offsetRgb);
        uint c2 = *(uint*)(rgbP + offsetRgb + 3);
        uint c3 = *(uint*)(rgbP + offsetRgb + 6);
        uint c4 = *(uint*)(rgbP + offsetRgb + 9);
        // swap R and B
        c1 = (c1 << 16) | (c1 & 0xFF00) | ((c1 >> 16) & 0xFF);
        c2 = (c2 << 16) | (c2 & 0xFF00) | ((c2 >> 16) & 0xFF);
        c3 = (c3 << 16) | (c3 & 0xFF00) | ((c3 >> 16) & 0xFF);
        c4 = (c4 << 16) | (c4 & 0xFF00) | ((c4 >> 16) & 0xFF);
        // set alpha to FF
        rgbaP[i] = c1 | 0xff000000;
        rgbaP[i + 1] = c2 | 0xff000000;
        rgbaP[i + 2] = c3 | 0xff000000;
        rgbaP[i + 3] = c4 | 0xff000000;
    }
}

或更简单的版本，每个迭代处理一个像素并且不使用unsafe，

void internalCvt(long pixelCount, byte[] bgr, byte[] rgba)
{
    long byteCount = pixelCount * 4;
    for (long i = 0, offsetBgr = 0; i < byteCount; i += 4, offsetRgb += 3)
    {
        // R
        rgba[i] = bgr[offsetBgr + 2];
        // G
        rgba[i + 1] = bgr[offsetBgr + 1];
        // B
        rgba[i + 2] = bgr[offsetBgr];
        // A
        rgba[i + 3] = 0xFF;
    }
}

如何将RGB摄像机图像转换为SharpDX的ARGB格式？

1 个答案: