rounding to the nearest zero, bitwise

Question

I just wonder how can i round to the nearest zero bitwise? Previously, I perform the long division using a loop. However, since the number always divided by a number power by 2. I decide to use bit shifting. So, I can get result like this:

12/4=3 13/4=3 14/4=3 15/4=3 16/4=4

can I do this by performing the long division like usual?

12>>2 13>>2

if I use this kind of bit shifting, are the behavior different for different compiler? how about rounding up? I am using visual c++ 2010 compiler and gcc. thx

I think it depends on a few factors, are you shifting signed or unsigned ints? What width? I found this link interesting for learning more about C and integers blog.regehr.org/archives/721 — Wayne Uroda
– Wayne Uroda, Commented Apr 9, 2013 at 3:35
In my experience if you write int x; int y; y = x/4; then any compiler worth a damn should convert that /4 into a bitshift for you (with optimisations turned on). — Wayne Uroda
– Wayne Uroda, Commented Apr 9, 2013 at 3:51

nneonneo · Accepted Answer · 2013-04-09 03:50:12Z

5

Bitwise shifts are equivalent to round-to-negative-infinity divisions by powers of two, meaning that the answer is never bigger than the unrounded value (so e.g. (-3) >> 1 is equal to -2).

For non-negative integers, this is equivalent to round-to-zero.

edited Apr 9, 2013 at 3:50

answered Apr 9, 2013 at 3:37

nneonneo

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4 Comments

Jo Skorsev Over a year ago

what do you mean by round to negative infinity? so, you are saying, the behavior produced by bitwise shifts isn't vary from one compiler to different compiler?

nneonneo Over a year ago

Provided you're using 2s-complement and arithmetic shifting for signed ints, it should be the same for all compilers.

Wayne Uroda Over a year ago

Round to negative infinity means if the result is say 2.5 or -2.5, the rounding always makes the number smaller (to the next smaller integer), so 2.x => 2 and -2.x => -3.

TylerH Over a year ago

Hi @nneonneo ...can you offer a source or spec sheet somewhere so I can read more about this? I'm getting into JS and using bitwise operators, and I'm trying to wrap my head around exactly how a bitwise operator will always truncate numbers after a decimal point. Yours is the first answer I've found that offers an explanation, and I'd like to know where this knowledge came from.

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