Convert from yuv 420 to image

Question

I have byte array with yuv420 data.

byte[] yuv420;//yuv data

How can I convert this to an Image?

I fo

Answer 1

One faster mode. Two mutiplication and two add less per pixel:

private static unsafe void YUV2RGBManaged(byte[] YUVData, byte[] RGBData, int width, int height)
{
    //returned pixel format is 2yuv - i.e. luminance, y, is represented for every pixel and the u and v are alternated
    //like this (where Cb = u , Cr = y)
    //Y0 Cb Y1 Cr Y2 Cb Y3 

    /*http://msdn.microsoft.com/en-us/library/ms893078.aspx
     * 
     C = 298 * (Y - 16) + 128
     D = U - 128
     E = V - 128
     R = clip(( C           + 409 * E) >> 8)
     G = clip(( C - 100 * D - 208 * E) >> 8)
     B = clip(( C + 516 * D          ) >> 8)

     * here are a whole bunch more formats for doing this...
     * http://stackoverflow.com/questions/3943779/converting-to-yuv-ycbcr-colour-space-many-versions
     */


    fixed(byte* pRGBs = RGBData, pYUVs = YUVData)
    {
        for (int r = 0; r < height; r++)
        {
            byte* pRGB = pRGBs + r * width * 3;
            byte* pYUV = pYUVs + r * width * 2;

            //process two pixels at a time
            for (int c = 0; c < width; c += 2)
            {
                int C1 = 298 * (pYUV[1] - 16) + 128;
                int C2 = 298 * (pYUV[3] - 16) + 128;
                int D = pYUV[2] - 128;
                int E = pYUV[0] - 128;

                int R1 = (C1 + 409 * E) >> 8;
                int G1 = (C1 - 100 * D - 208 * E) >> 8;
                int B1 = (C1 + 516 * D) >> 8;

                int R2 = (C2 + 409 * E) >> 8;
                int G2 = (C2 - 100 * D - 208 * E) >> 8;
                int B2 = (298 * C2 + 516 * D) >> 8;

                //check for overflow
                //unsurprisingly this takes the bulk of the time.
                pRGB[0] = (byte)(R1 < 0 ? 0 : R1 > 255 ? 255 : R1);
                pRGB[1] = (byte)(G1 < 0 ? 0 : G1 > 255 ? 255 : G1);
                pRGB[2] = (byte)(B1 < 0 ? 0 : B1 > 255 ? 255 : B1);

                pRGB[3] = (byte)(R2 < 0 ? 0 : R2 > 255 ? 255 : R2);
                pRGB[4] = (byte)(G2 < 0 ? 0 : G2 > 255 ? 255 : G2);
                pRGB[5] = (byte)(B2 < 0 ? 0 : B2 > 255 ? 255 : B2);

                pRGB += 6;
                pYUV += 4;
            }
        }
    }
}