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Home/ Questions/Q 261659
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Editorial Team
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Editorial Team
Asked: May 11, 20262026-05-11T22:27:02+00:00 2026-05-11T22:27:02+00:00

Possible Duplicate: How to represent 18bit color depth to 16bit color depth? I’m porting

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Possible Duplicate:
How to represent 18bit color depth to 16bit color depth?

I’m porting a software that build from 16-bit color depth device to 18-bit color depth device? How can I represent the 18-bit color depth? Thanks.

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  1. Editorial Team
    Editorial Team
    2026-05-11T22:27:02+00:00Added an answer on May 11, 2026 at 10:27 pm

    NOTE: This is just an edit of my answer to your previous question, since the two are so similar – you just have to reverse the process.


    You will first of all have to access each of the colour components (i.e. extract the R value, the G value, and the B value). The way to do this will depend totally on the way that the colour is stored in memory. If it is stored as RGB, with 5-6-5 bits for the components, you could use something like this:

    blueComponent  = (colour & 0x3F);
    greenComponent = (colour >>  6) & 0x3F;
    redComponent   = (colour >> 12) & 0x3F;
    

    This will extract the colour components for you, then you can use the method outlined above to convert each of the components (I presume 18-bit will use 6 bits per component):

    blueComponent  = (blueComponent  / 64.0) * 32;
    //greenComponent = (greenComponent / 64.0) * 64;    //not needed
    redComponent   = (redComponent   / 64.0) * 32;
    

    Note that with the above code, it is important that you use 64.0, or some other method that will ensure that your program represents the number as a floating point number. Then to combine the components into your 18-bit colour, use:

    colour = (redComponent << 11) | (greenComponent << 5) | blueComponent;
    
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