Today I've written a little program that classifies and counts files based on the integer part of log2($file_size). So I would get these classes: 
0 <= SIZE < 1
1 <= SIZE < 2
2 <= SIZE < 4
4 <= SIZE < 8
and so on...

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For that I would like to write this: 
$size_class = $size ? int( log($size) / log(2) ) : -1;
# red light on, possible error caused by floating point arithmetic

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Now let's forget about the special case of $size == 0. My input can only be positive integers (because these are file sizes), so I suppose the error coming from floating point arithmetic can bite me only in the case when $size is a power of 2 (in the expression $size = 2**$x the exponent is a natural number). Is this true?

But let's take it further. I've written a little test to check whether I will be bitten for all the powers of 2 that fits in my machine's floating point represantation: 

bash $ diff <( perl -E 'say int( log(2**$_) / log(2) ) for 1 .. 2**10'
+ ) <( seq 1 $(( 2**10 )) )

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The output is this: 
1024c1024
< inf
---
> 1024

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So there is not a single power of 2 (under 2**1024) that would be misclassified at least on my machine.

So my question is: Is there a theoretical explanation (for this restricted case of floating point aritmetic) why it can't be problematic or is it just by accident?

Please shed some light on this issue.

ps: Sorry, I know this is probably not Perl specific, but my question arose because there is no log2() function in Perl. (For example with a properly setup lookup table I could be able to do this classification with integer arithmetic only, but that would be a pain instead of this little snippet.)

In reply to May I be bitten by floating point arithmetic in the following restricted case? by rubasov

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