in reply to Why the Integer part of this calculation is 102?

$ perl -e 'printf "%.20f\n", 42435.3408/411.9936'
102.99999999999998578915

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This is smaller than 103, so int rounds down to 102. You'll have to use proper rounding.

Perl 6 - links to (nearly) everything that is Perl 6.

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Re^2: Why the Integer part of this calculation is 102?
by alokam (Initiate) on Mar 22, 2010 at 21:46 UTC
    Thanks for the quick response on my issue! I am still confused, sorry! But wouldn't we get exactly 42435.3408 when we multiply 411.9936 * 103? Is the internal representation of 42435.3408 is different in perl?
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      Both 3408/10000 and 9936/10000 are periodic numbers in binary, just like 1/3 is a periodic number in decimal. It would take infinite storage to store them as floats, so they're not being stored exactly.

      You can avoid the problem by using sprintf("%.0f", $x) (rounds) instead of int($x) (truncates).

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      Others have answered you already; I just want to add that in Perl 6 your code does what you expect. It achieves that by storing decimal numbers as fractions/rationals internally. 
      $ 
perl6
> say (42435.3408).perl
26522088/625
> say (42435.3408).WHAT
Rat()
> say 42435.3408 / 411.9936
103
> printf "%.20f\n", 42435.3408 / 411.9936
103.00000000000000000000

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      Perl 6 - links to (nearly) everything that is Perl 6.
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[d/l]
        Thank you all for your quick response on my issue. Greatly appreciate your help. -Anju
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