Color Conversion

This is a discussion on Color Conversion within the C Programming forums, part of the General Programming Boards category; I use the following to convert from ARGB1555 to ARGB8888 Code: TempColor = ((pColor->Value.argb.B & 0xF8) << 7) | ((pColor->Value.argb.G ...

  1. #1
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    Color Conversion

    I use the following to convert from ARGB1555 to ARGB8888

    Code:
                TempColor = ((pColor->Value.argb.B & 0xF8) << 7) |
                            ((pColor->Value.argb.G & 0xF8) << 2) |
                            ((pColor->Value.argb.R & 0xF8) >> 3) |
                            ((pColor->Value.argb.A & 0x80) << 8);
    Can you anyone tell me equivalent code for converting ARGB8888 to ARGB1555

    Thanks in advance

  2. #2
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    It looks like you are combining the B,G,R,A values into an integer with the bits having the following meanings:

    abbbbbgggggrrrrr

    So to split it out into separate components
    Code:
    pColor->Value.argb.A = TempColor >> 8 & 0x80;
    pColor->Value.argb.R = TempColor << 3 & 0xF8;
    pColor->Value.argb.G = TempColor >> 2 & 0xF8;
    pColor->Value.argb.B = TempColor >> 7 & 0xF8;

  3. #3
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    Quote Originally Posted by nonoob View Post
    It looks like you are combining the B,G,R,A values into an integer with the bits having the following meanings:

    abbbbbgggggrrrrr

    So to split it out into separate components
    Code:
    pColor->Value.argb.A = TempColor >> 8 & 0x80;
    pColor->Value.argb.R = TempColor << 3 & 0xF8;
    pColor->Value.argb.G = TempColor >> 2 & 0xF8;
    pColor->Value.argb.B = TempColor >> 7 & 0xF8;
    So the above one gives ARGB values in ARGB8888 format, where TempColor is in APRG1555 ?

  4. #4
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    That code isn't quite right. When converting the 5-bit channel values to 8-bit quantities a value of 0x1F (31) should convert to 0xFF (255), not 0xF8 (248). The rest between 0 and 0x1F should smoothly map to values from 0 to 0xFF. The way to do this is to repeat the bit pattern.
    Taking 0x1A for example, and dropping down to binary:
    Code:
    0x1A:
    11010
         11010 (same as above again)
    11010110   (taking first 8 bits)
    ==> 0xD6
    Thus our answer is that 0x1A gets converted into 0xD6, not 0xD0 which you would get without repeating the bit pattern.
    Now that seems like more work, however it can be done very fast by using some clever tricks to do the bit repeating on all 3 RGB channels at once.
    Note that for the alpha channel we're converting from a 1-bit quantity to an 8-bit quantity. 0 maps to 0 and 1 maps to 255, that's it. Since we have to repeat the bit pattern 8 times, it's actually quicker to achieve that through a multiply, or if you prefer, a shift plus a 2 byte lookup table.
    See the below code which is optimised to about as fast as these correct conversions can possibly be done.

    When you convert back from 8-bit RGB channels to 5-bit, you just drop the extra bits.
    Code:
    unsigned int ARGB1555toARGB8888(unsigned short c)
    {
        const unsigned int a = c&0x8000, r = c&0x7C00, g = c&0x03E0, b = c&0x1F;
        const unsigned int rgb = (r << 9) | (g << 6) | (b << 3);
        return (a*0x1FE00) | rgb | ((rgb >> 5) & 0x070707);
    }
    unsigned short ARGB8888toARGB1555(unsigned int c)
    {
        return (unsigned short)(((c>>16)&0x8000 | (c>>9)&0x7C00 | (c>>6)&0x03E0 | (c>>3)&0x1F));
    }
    You'll need direct access to the unsigned short and unsigned int in each case, so if you can't work that out, post the definitions of the data types for "Value" etc.
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