Anonymous Monk has asked for the wisdom of the Perl Monks concerning the following question:

Maybe I am missing something very obvious, but why is the following behavior going on in Perl (5.14.2)? 
3 % 3 # returns 0
log (1000) / log (10) # returns 3
(log(1000) / log(10)) % 3 # returns 3

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Re: Strange log behavior?
by ikegami (Patriarch) on Sep 07, 2013 at 16:42 UTC 
    >perl -E"say sprintf '%.20g', log (1000) / log (10)"
2.9999999999999996

>perl -E"say int( log (1000) / log (10) )"
2

>perl -E"say log (1000) / log (10) % 3"
2

>perl -E"say 2 % 3"
2

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    Always use a tolerance when comparing floating point numbers or converting them to integers.

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      What is the correct way to round the floating point number to 3?
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        sprintf seems to do the right thing.

        perl -E '
say sprintf q{%.0f}, $_ for
   2.99999,
   2.49999,
   3,
  -2.99999,
  -2.49999,
  5279.8012;'
3
2
3
-3
-2
5280
$

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        I hope this is helpful.

        Cheers,

        JohnGG

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Re: Strange log behavior?
by Eily (Monsignor) on Sep 07, 2013 at 16:51 UTC

    You probably wrote something like : print (log(1000) / log(10)) % 3 Which, has stated in perlop, actually means (print (log(1000 / log(10))) % 3.

    So you print (log 100)/(log 10) (and the result of this operation is rounded to 3), and then do a modulo operation on the output of print, which doesn't make much sense.

    Edit: if that is what happened, use warnings would have given you the answer

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