Bit twiddling for checking whether a number is in particular range

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I found some interesting bit twiddling in "source\common\unicode\utf.h" file of ICU library (International Components for Unicode). The bit twiddling is intended for checking whether a number is in a particular range.

// Is a code point in a range of U+d800..U+dbff?
#define U_IS_LEAD(c) (((c)&0xfffffc00)==0xd800)

I have figured out the magic number (0xfffffc00) come from:

MagicNumber = 0xffffffff - (HighBound - LowBound)

However, I also found that the formula doesn't apply to every arbitrary range. Does somebody here know in what circumstance the formula works?

Is there another bit twiddling for checking whether a number is in particular range?

c++
c
algorithm
bit-manipulation
asked on Stack Overflow Jan 1, 2011 by Astaroth • edited Apr 10, 2019 by David Jones

3 Answers

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For these tricks to apply, the numbers must have some common features in their binary representation.

0xD800 == 0b1101_1000_0000_0000
0xDBFF == 0b1101_1011_1111_1111

What this test really does is to mask out the lower ten bits. This is usually written as

onlyHighBits = x & ~0x03FF

After this operation ("and not") the lower ten bits of onlyHighBits are guaranteed to be zero. That means that if this number equals the lower range of the interval now, it has been somewhere in the interval before.

This trick works in all cases where the lower and the higher limit of the interval start with the same digits in binary, and at some point the lower limit has only zeroes while the higher limit has only ones. In your example this is at the tenth position from the right.

answered on Stack Overflow Jan 1, 2011 by Roland Illig • edited May 26, 2015 by Roland Illig
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If you do not have 2^x boundaries type might use the following trick:

if x >= 0 and x < N you can check both by:

  if Longword( x ) < Longword( N ) then ...

This works due to the fact that negative numbers in signed numbers correspond to the largest numbers in unsigned datatypes.

You could extend this (when range checking is DISABLED) to:

  if Longword( x - A ) < Longword ( ( B - A ) ) then ...

Now you have both tests (range [ A, B >) in a SUB and a CMP plus a single Jcc, assuming (B - A ) is precalculated.

I only use these kind of optimizations when really needed; eg they tend to make your code less readable and it only shaves off a few clock cycles per test.

Note to C like language readers: Longword is Delphi's unsigned 32bit datatype.

answered on Stack Overflow Jan 1, 2011 by Ritsaert Hornstra • edited Jan 1, 2011 by Ritsaert Hornstra
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The formula works whenever the range you are looking for starts at a multiple of a power of 2 (that is, 1 or more bits at the low end of the binary form of the number ends in 0) and the size of the range is 2^n-1 (that is, low&high == low and low|high == high).

answered on Stack Overflow Jan 1, 2011 by Vatine

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