From: Alexander Schofield <pcs3@...>
Date: 2002-02-08T06:24:15+09:00
Subject: Re: Why is one slower than the other?

In B you're instantiating a new Regexp on each iteration.

Kirk Haines wrote:
> 
> Program A:
> 
> z = 0
> t = "0123456789abcdefghijklmnopqrstuvwxyz"
> (0..999999).each {
>   z += 1 if t =~ /abc/
> }
> 
> Program B:
> 
> z = 0
> t = "0123456789abcdefghijklmnopqrstuvwxyz"
> rexp = Regexp.new('abc');
> (0..999999).each {
>   z += 1 if rexp.match t
> }
> 
> B has a runtime that is about 1.7 times A.  Why?  This is on Ruby 1.6.6 on
> a Linux box.
> 
> Kirk Haines
> 
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-- 
Alexander Schofield