in reply to Fibonacci numbers

Here's the most simple program to do it. You can test it. 
$a=0;
$b=1;
while($a <=200)
    {print("$a\n");
     ($a,$b)=($b,$a+$b);
    }

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Re: Re: Fibonacci numbers(Again)
by Anonymous Monk on Oct 02, 2001 at 23:48 UTC
    How does this statement 
    ($a,$b)=($a,$a+$b);

    [download]
    does the trick? I couldn't understand...can somebody explain?
[reply]
[d/l]

      Perhaps you can't see it because you've got the statement wrong. It should be

      ($a,$b)=($b,$a+$b);

      [download]

      It's a simple list assignment. $a gets the value of $b and $b gets $a+$b.

      Incidently, it can be a bad idea to use $a and $b in Perl scripts as they are "special" variables because of their use in sorting.

      Blessed Be
      The Pixel

[reply]
[d/l]
      Dear Anonymous, It's very simple. In every iteration the $a receives the earlier value of $b...that's the trick. Bye the way, u can see my ealier attempt, which had one serious mistake... that can be rewritten to give the desired result. See: 
      $a=1;
$b=1;

while($a <=200)
    {$c=$a+$b;
     print("$a\n$b\n$c\n");
    $a=$b+$c;
    $b=$c+$a;
    }

      [download]
      Thanks.
[reply]
[d/l]

      Oh, for goodness sake. This is homework, right?

      For the answer to this latest question, see the first answer in the thread...

      (Alternatively, read what you're quoting: it might make more sense then.)

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