YUV420 轉 RGB565

R =  1.164 * (Y-16) +                 + 1.596 * (V-128)           final : 5 bit
G =  1.164 * (Y-16) - 0.391 * (U-128) -  0.813 * (V-128)           final : 6 bit
B =  1.164 * (Y-16) + 2.018 * (U-128)                             final : 5 bit

R.G.B Range : (Y.U.V are 8 bits):

max : B : 1.164 * (255-16) + 2.018 * (255-128) = 534.482
min : G : 1.164 * (0 - 16) - 0.391 * (255-128) - 0.813 * (255-128) = -171.832

所以 8 bit YUV 轉成 RGB 後，最高只有到
take advantage of the final 5-6-5 bits.
source not need to be 8 bits.

R =  ( 1.164 * (Y-16) +                 + 1.596 * (V-128) )/8          final : 5 bit
G =  ( 1.164 * (Y-16) - 0.391 * (U-128) - 0.813 *(V-128)  )/4         final : 6 bit
B =  ( 1.164 * (Y-16) + 2.018 * (U-128)                   )/8          final : 5 bit

變成：

R =  0.1455 * (Y-16) +                   + 0.1995 * (V-128)           final : 5 bit
G =   0.291 * (Y-16) - 0.09775 * (U-128) - 0.04575 *(V-128)           final : 6 bit
B =  0.1455 * (Y-16) + 0.25225 * (U-128)                              final : 5 bit

目標是希望..

word rgb565 = ( RY[Y]+RV[V] ) | ( GY[Y]+GU[U]+GV[V] ) | ( BY[Y]+BU[U] )

因為是 YUV420 ，最好是..(U.V矩陣 的HighWord和LowWord一樣)

DWORD rgb565x2 =( (RYL[Y1]|RYH[Y2])+RV[V] )       |
             ( (GYL[Y1]|GYG[Y2])+GU[U]+GV[V] ) |
             ( (BYL[Y1]|BYH[Y2])+BU[U] )

問題：如何確保 + 後不會16 bit overfload ?

if possible 再共用下一個scanline (因為是 YUV420)..

RP=RV[v];
GP=GU[U]+GV[V];
BP=BU[U];

DWORD rgb565x2linea =( (RYL[Y1a]|RYH[Y2a])+RP )       |
                 ( (GYL[Y1a]|GYG[Y2a])+GP ) |
                 ( (BYL[Y1a]|BYH[Y2a])+BP );
DWORD rgb565x2lineb =( (RYL[Y1b]|RYH[Y2b])+RP )       |
                 ( (GYL[Y1b]|GYG[Y2b])+GP ) |
                 ( (BYL[Y1b]|BYH[Y2b])+BP );

但是好像無法保證不會overflow (saturation)。
回到：

word rgb565 = ( RY[Y]+RV[V] ) | ( GY[Y]+GU[U]+GV[V] ) | ( BY[Y]+BU[U] )

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ref http://msdn.microsoft.com/en-us/library/ms867704.aspx