Solved

What is the use of the shift operator <<?

Posted on 2004-09-05
3
461 Views
Last Modified: 2010-04-01
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
HCK
0
Comment
Question by:hengck23
[X]
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
  • 2
3 Comments
 
LVL 19

Accepted Solution

by:
drichards earned 205 total points
ID: 11987404
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.
0
 
LVL 11

Assisted Solution

by:avizit
avizit earned 20 total points
ID: 11987474
also "shift" instructions in a processor are expected to be faster than a multiply instruction
0
 
LVL 19

Assisted Solution

by:drichards
drichards earned 205 total points
ID: 11987612
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.
0

Featured Post

On Demand Webinar - Networking for the Cloud Era

This webinar discusses:
-Common barriers companies experience when moving to the cloud
-How SD-WAN changes the way we look at networks
-Best practices customers should employ moving forward with cloud migration
-What happens behind the scenes of SteelConnect’s one-click button

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Suggested Solutions

Title # Comments Views Activity
Better understanding on C++ Class serialization and formats 9 103
Copy output image from TWindowsMediaPlayer 6 67
what type of pointer is this pointer? 3 31
visual C++ 1 41
Often, when implementing a feature, you won't know how certain events should be handled at the point where they occur and you'd rather defer to the user of your function or class. For example, a XML parser will extract a tag from the source code, wh…
What is C++ STL?: STL stands for Standard Template Library and is a part of standard C++ libraries. It contains many useful data structures (containers) and algorithms, which can spare you a lot of the time. Today we will look at the STL Vector. …
The viewer will learn additional member functions of the vector class. Specifically, the capacity and swap member functions will be introduced.
The viewer will learn how to clear a vector as well as how to detect empty vectors in C++.

734 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question