From: Daniel Carrera <dcarrera@...>
Date: 2003-07-24T08:48:14+09:00
Subject: Re: "regex literal in condition"

Thanks for the explanation, but I'm not quite following.
I now understand what a string literal is, but I don't quite get the 
problem at the if.

HTH wrote:
[snip]
> Notice how if you do
> 
> re = /\<\!-{19} ending left_colum -{11}\>/
> while gets
>     break if re 
> end   
> 
> then "gets" is only called once because any object instance turns true
> in a boolean condition.

Uhm... Let's see if I understand what you are saying.  "re" here is an 
instance of the class Regexp.  Since it's an object instance, it is 
interpreted as "true" by the if.  Did I understand that right?

What I don't follow is:
  -  Why would that be the case?
  -  How is a non-literal any less of an object instance than a literal.

I guess that the best solution is to use ($_ =~ re).  But I'd still like 
to understand why Ruby behaves this way.

Here is another example of weird example:

$ irb
>> re = /hi/
>> $_ = "hello"
>> puts "yes" if ($_ =~ re)
=> nil
>> puts "yes" if re
yes
=> nil

Thanks for the help.
-- 
Daniel Carrera    | OpenPGP fingerprint:
Mathematics Dept. | 6643 8C8B 3522 66CB D16C D779 2FDD 7DAC 9AF7 7A88
UMD, College Park | http://www.math.umd.edu/~dcarrera/pgp.html