Re: Re: Fibonacci numbers(Again)

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Re: Re: Re: Fibonacci numbers(Again) by pixel (Scribe) on Oct 03, 2001 at 14:02 UTC
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]
Re: Re: Re: Fibonacci numbers(Again) by saiful (Novice) on Oct 03, 2001 at 13:45 UTC
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]
Re: Re: Re: Fibonacci numbers(Again) by tommyw (Hermit) on Oct 03, 2001 at 02:49 UTC
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.)	[reply]