in reply to RE: RE: RE: (Ovid) Re: General quesion
in thread General quesion

Yes it does automatically takethe integer portion when using it for an array index. Using int() you will also simply get the integer portion, not a rounded number, so you using the int() is sort of a noop. To round, you could do this:

my $foo = sprintf "%.0f", rand(100);

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But, to illustrate the answer to your question:

perl -wle '@r = (1,2,3); print $r[$t = rand(3)]; print "$t"';
2
1.28954588016495

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Cheers,
KM

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RE: RE: RE: RE: RE: (Ovid) Re: General quesion
by tye (Sage) on Aug 08, 2000 at 23:27 UTC

    Hmm. I'm not sure what you mean but I think you have made an incorrect assumption. Perl's int() just rounds down. It does not (necessarilly) convert the number into a long or unsigned long. You might want to read my previous node on Perl and floating point.

            - tye (but my friends call me "Tye")
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[d/l]
[select]
      No wrong assumption here. I said int() get's the integer portion of the number, which echoes the FAQ:

      # perldoc -f int
=item int EXPR

=item int

Returns the integer portion of EXPR. etc...
      Reread my post, I make no mention of int() rounding. And you are incorrect, Perl's int() doesn't round down, it simply gives you the integer portion. 
      # perl -wle 'print int(1.66)'

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      But, I do see where my wording may have been confuzzling, so I changed it a little.

      Cheers,
      KM

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[d/l]

        Yes it does automatically round the number down (which is simply taking the integer portion)

        Perl's int() doesn't round down, it simply gives you the integer portion

        Um, you might want to make up your mind on what "rounding down" means. :)

                - tye (but my friends call me "Tye")
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