Perl optimisation help

[david thornton]

Rick Delaney wrote:

>On Fri, Jun 09, 2006 at 12:47:14AM +0300, Peter wrote:
>  
>
>>On Thu, 8 Jun 2006, Lennart Sorensen wrote:
>>
>>    
>>
>>>On Thu, Jun 08, 2006 at 11:07:45PM +0300, Peter wrote:
>>>      
>>>
>>>>I did it with a hash and it is 5 times faster. Still I would like to
>>>>know what the fastest way to do $var=$var.$add; is.
>>>>        
>>>>
>>>Perl don't have someting icky like $var .= $more; does it?
>>>      
>>>
>>Not that I know of it.
>>    
>>
>
>It does.  If you don't like the perlmonks suggestion you might want to
>try the local Toronto Perl Mongers mailing list, tpm-BnhL/mGhxvQEbZ0PF+XxCw at public.gmane.org
>
>  
>
>>>How do you do it with a hash?
>>>      
>>>
>>%hash=();
>>$idx=0;
>>
>>cond loop {
>>  $hash{$idx++}=$line;
>>}
>>
>>$res='';
>>$i=0;
>>while($i<$idx) {
>>  $res = $res.$hash{$i++};
>>}
>>
>>this is fast, and could be faster by adding buckets. I think that Perl 
>>measures the length of the string and caches it. If the next access is 
>>'soon enough' the length must not be calculated again. The speedup 
>>factor is between 6 and 13 times (!) vs. using '.' standalone.
>>    
>>
>
>Your problem is certainly not with '.' since you are still using it.
>More likely you have nested loops or something but it is really
>difficult to tell what you're trying to do with your pseudocode.
>
>If all you want is all input in one variable then there are many ways to
>do so directly.  One would be
>
>    $var = join "", <INPUT>;
>
>Faster would be
>
>    { local $/; $var = <INPUT>; }
>
>
>  
>
I realize there's not much to it, but can you walk us through "local $/" 
? I love brief code, just it's sometimes hard to understand.

David
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