Subject: [ruby-talk:7455] Re: question: 'A'..'k'
From: Mark Slagell <ms iastate.edu>
Date: Sun, 17 Dec 2000 03:38:24 +0900
References: 7449
Mathieu Bouchard wrote:
>
> k=0; ("A".."k").each{|x|k+=1}; k
>
> I expected 11, I got 26. Why is that?
>
> matju
k=0; (?A..?k).each{|x|k+=1}; k
... yields 43, and I think that makes sense. How does it work for
strings?