in reply to $var == 1 fails when $var = 1

floating-point numbers have rounding errors and our decimal system isn't too compatible to binary representation

DB<8> printf '%.20f', 0.688 + 0.289 + 0.023
0.99999999999999989000

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so $var != 1

Perl's DWIM is sometimes trying to round the string representation, that's why eq seemingly "works".

Rule of thumb, add integers and correct afterwards to the right magnitude.

  DB<15> p 1 == (688 + 289 + 23) /1000
1

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see also Humans have too many fingers

Cheers Rolf
(addicted to the Perl Programming Language :)
Wikisyntax for the Monastery FootballPerl is like chess, only without the dice

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Re^2: $var == 1 fails when $var = 1
by mjlush (Novice) on Sep 20, 2018 at 15:23 UTC
    Why does the order of addition matter? 
    DB<8> printf '%.20f', 0.688
0.68799999999999994493
DB<9> printf '%.20f', 0.688 + 0.023
0.71099999999999996536
DB<10> printf '%.20f', 0.688 + 0.023  + 0.289
1.00000000000000000000
DB<11> printf '%.20f', 0.688 + 0.289 + 0.023
0.99999999999999988898
    PS and thanks for the prompt answer!
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      Because you are accumulating different rounding errors and (most likely) a shifting mantissa changes the last bit to be rounded.

      update

      Rounding is often not very intuitive, try to avoid if possible.

      If you want to track the details you need to look into the binary representation and the underlying C lib.

      Cheers Rolf
      (addicted to the Perl Programming Language :)
      Wikisyntax for the Monastery FootballPerl is like chess, only without the dice

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        That sort of makes sense :-) Is there a module that covers over these sort of cracks? It would seem to be a routine problem.
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