[ruby-core:24832] Re: Proposal: match? with no backrefs

If you inspect object allocations, you will see that there are more 
object allocations regardless of use of $1. I would like a way to check 
for regular expression match without creating any objects at all.

-- Yehuda

NARUSE, Yui wrote:
> Kirk Haines wrote:
>    
>> On Tue, Aug 4, 2009 at 9:37 PM, NARUSE, Yui<[e-mail]>  wrote:
>>      
>>> 2009/08/03 9:04, Yehuda Katz wrote:
>>>        
>>>> In both of those cases, $1 is set.
>>>>          
>>> $1 is virtual variable.
>>> The value of $1 is created when it is called.
>>>        
>> I think what Yehuda is getting at, though, is that:
>>
>> 'abcdefghijklmnopqrstuvwxyz' =~ /nop/
>>
>> is quite a bit faster than:
>>
>> 'abcdefghijklmnopqrstuvwxyz' =~ /n(op)/
>>      
>
> If you don't use $1, /n(?:op)/ seems better.
>
>    
>> So it would be nice to be able to:
>> r = /n(op)/
>> 'abcdefghijklmnopqrstuvwxyz'.match?(r)
>>      
>
> r = /n(?:op)/
> r =~ 'abcdefghijklmnopqrstuvwxyz'
> is not enough?
>
>