From: Rob Biedenharn <Rob@...>
Date: 2009-07-17T04:30:41+09:00
Subject: Re: case statement puzzle

On Jul 16, 2009, at 12:49 PM, Tom Cloyd wrote:

> I'm missing something here, and cannot see the problem:
>
> x='1'
> (1..5).include? x.to_i  # => true
>
> But...
>
> x='1'
> case
> when x =='0'
> puts '0'
> when (1..5).include? x.to_i
> puts '1'
> end
>
> ...won't even compile. Can someone tell me why? (and maybe how to  
> fix it...)
>
> What I'm having to do is this, which works:
>
> ...
> when (1,,5).to_a & [x.to_i].length > 0
> ...
>
> but it seems over-wrought.
>
> t.
> -- 
> ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
> Tom Cloyd, MS MA, LMHC - Private practice Psychotherapist
> Bellingham, Washington, U.S.A: (360) 920-1226
> << tc@tomcloyd.com >> (email)
> << TomCloyd.com >> (website) << sleightmind.wordpress.com >> (mental  
> health weblog)
> ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


x='1'
case
when x =='0'
   puts '0'
when (1..5).include?(x.to_i)
   puts '1'
end

If you don't leave the parentheses off of the .include? method it will  
work. The better answers boil down to "learn about the case-equality  
method ===", but using parentheses will at least get around the  
ambiguous parsing.

You should probably also consider that the value of the entire case  
statement, as written, will be nil since that is the "return value" of  
the puts method as well as the  value if none of the 'when' clauses  
match.

-Rob

Rob Biedenharn		http://agileconsultingllc.com
Rob@AgileConsultingLLC.com


P.S. For completeness:

x = '1'
case x
when '0'
   puts '0'
when '1'..'5'
   puts '1'
end