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

My turn to go Aargh!!! I just wrote pretty much this same code and I missed the for (1..8) optimization.

Of course, I have an excuse since my code actually accepts the size for the password and if they go with a real large password, I would have to wait for the list to be created. Yeah, that's my excuse and I'm sticking to it. :-) Yes, I know it's a weak excuse, but it was the best I could come up with.

One question though. When using a random number it returns a decimal. Does using it as an array index automagically int() it? 'Cause in my code, I explicitly call int around my rand call.

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RE: RE: (Ovid) Re: General quesion
by Ovid (Cardinal) on Aug 08, 2000 at 04:43 UTC
    When using a random number it returns a decimal. Does using it as an array index automagically int() it?

    Take a look at the scalar we're creating: 

    $myarray[rand(@myarray)];

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    You'll notice that rand is being passed @myarray. Since rand is expecting a scalar, @myarray is interpreted in a scalar context. Therefore, it returns the length of the array (52 in this case), which will cause rand to generate a random number from 0 to 51, which fits the array index quite nicely. Perhaps this would have been more clear: 
    $myarray[rand($#myarray)];

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    Cheers,
    Ovid
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      But rand($#myarray) returns 0..50 not 0..51. $#array in a scalar context is always one less than @array in a scalar context.

      So, the idiom is: 

      $random_item = $array[rand @array];

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      Just cut-n-paste that!

      -- Randal L. Schwartz, Perl hacker

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      My question wasn't clear. I know that rand() returns a number between 0 and it's argument. But my point was, it returns a decimal; i.e. 3.14159, 25.70098, .00001, etc. Does perl automatically round the number down(or take the integer portion)? For eaxample, here is my code:

          my $lRandomPassword = '';
    my $lMaxLength = 10;
    my $length;
    my $lRand;
    my @lKeySpace = ('A'..'Z', 'a..z', 0..9);
    my $lKeySpaceSize = scalar(@lKeySpace);
    for ($length = 0; $length < $lMaxLength; $length++) {
        $lRand = int(rand($lKeySpaceSize));
        $lRandomPassword .= $lKeySpace[$lRand];
    };
    $lRandomPassword;

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      Yes, I know the code is overly verbose, but the important question for me is, could I safely get rid of the int() call? Does perl take the integer portion of an array index?

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        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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