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What is the use of the shift operator <<?

Dear All,

I am writing the code for RGB to gray level conversion.
The NTSC standard for luminance:

I = 0.299r + 0.587g + 0.114b

While searching the web, i came across the code that does the task:

#define RGB_SCALE 14
#define cR (int)(0.299*(1 << RGB_SCALE) + 0.5)
#define cG (int)(0.587*(1 << RGB_SCALE) + 0.5)
#define cB ((1 << RGB_SCALE) - cR - cG)

//I is greyLevel output
//r is red value
//g is green value
//b is blue value

I = cR*r+cG*g+cB*b;
I = (((I) + (1 << ((RGB_SCALE)-1))) >> (RGB_SCALE));

What is the purpose of the bitwise operations involving RGB_SCALE, especially for the last line of code?

Thank you.

With Best Regards
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3 Solutions
In the #define's, the shifts are just to scale the values up by 2^14in order to use integer values rather than floats.  Note that cR and cG are cast back to int and cB uses integer arithmetic and does not need to be converted back.

In the final line, the value is shifted back (scaled down by 2^14) to get back to a normal integer grayscale value.  Each of RGB is an 8-bit number, and the coefficients in the grayscale calculation add up to 1, so the grayscale value is also in the range 0-256 (8 bits).  The '+ (1 << ((RGB_SCALE)-1))' part is just to round to what will be the LSB after the final right shift.

In short, this algorithmj just multiplies the original RGB values by a big number so integer math can be used to compute the grayscale value.  Then after the calculation, the big number is divided back out.  The '+ 0.5' and the '+ (2^13)' just round results.
also "shift" instructions in a processor are expected to be faster than a multiply instruction
The way this code is written, there is only one shift actually done in the executable code - the final right shift.  cR, cG, cB, and (1 << ((RGB_SCALE)-1)) are computed by the preprocessor and put into the code as constants.

And you would expect a shift to be much faster than a divide.

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