Problem with understanding masking operation

Thread Starter

zulfi100

Joined Jun 7, 2012
656
Hi,
I have a problem in understanding the following masking operation:

m & n = n if m + 1 = 2 ^ (1+log m) and 0 <= n <= m


where & is the bitwise AND operator and m denotes a bitmask 0xff…f.


I have attached the research paper. Its on page 3.


Some body please guide me.

Zulfi.
 

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Thread Starter

zulfi100

Joined Jun 7, 2012
656
Hi,
Okay I got some idea. What ever be the value of n, we get it back due to masking operation.
But can some body explain the part: m + 1 = 2 ^ (1+log m) by giving some example
Zulfi.
 

panic mode

Joined Oct 10, 2011
1,886
if n is less than m and m is a "block of set bits" then m & n = n
block of set bits means all "1" are grouped and on the right, for example m = 7 = binary 00000111
adding one to m would make it one of the binary powers: m+1 = 8 = binary 0001000
 

WBahn

Joined Mar 31, 2012
26,398
Hi,
Okay I got some idea. What ever be the value of n, we get it back due to masking operation.
But can some body explain the part: m + 1 = 2 ^ (1+log m) by giving some example
Zulfi.
What is the log operation? Is it base 2? Does it return the real-values logarithm, or does it floor or ceiling it?
 

Thread Starter

zulfi100

Joined Jun 7, 2012
656
What is the log operation? Is it base 2? Does it return the real-values logarithm, or does it floor or ceiling it?
Nothing like this mentioned in the paper. I thing we can consider it as returning the actual value. We can try both ceil and floor operations on it to see their impact. But you are right this is a confusion and its not well expressed.

Zulfi.
 

WBahn

Joined Mar 31, 2012
26,398
Hi,
Okay I got some idea. What ever be the value of n, we get it back due to masking operation.
But can some body explain the part: m + 1 = 2 ^ (1+log m) by giving some example
Zulfi.
Notice that the paper gives a bib reference. Go look at the paper cited.
 
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