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comp.lang.ruby

Re: nil question

e

1/30/2005 4:06:00 PM

> Lähettäjä: "David A. Black" <dblack@wobblini.net>
> Aihe: Re: nil question
>
> Hi --
>
> On Mon, 31 Jan 2005, Christian Neukirchen wrote:
>
> > "David A. Black" <dblack@wobblini.net> writes:
> >
> >> Hi --
> >>
> >> On Sun, 30 Jan 2005, Christian Neukirchen wrote:
> >>
> >>> "William James" <w_a_x_man@yahoo.com> writes:
> >>>
> >>>> Sam Roberts wrote
> >>>>> In ruby, zero and empty strings are true
> >>>>
> >>>> Since 0 is true, you should be able to do this in Ruby:
> >>>>
> >>>> puts "yes" if -5 < x < 9
> >>>>
> >>>> The phrase '-5 < x' should yield the value of x instead of true.
> >>>> That's the way it actually works in the Icon programming language.
> >>>> But we have to use the klunky
> >>>>
> >>>> puts "yes" if -5 < x and x < 9
> >>>>
> >>>
> >>> Erm, say, x is -16:
> >>>
> >>> (-5 < x) < 9
> >>> (-5 < -16) < 9
> >>> -5 < 9
> >>> -5
> >>>
> >>> -5 is true, probably not what you want.
> >>
> >> But -5 < -16 is not true, so it wouldn't get that far. (I assume
> >> William means it should return x if the expression is true, false
> >> otherwise.)
> >
> > So false is bigger than 9? Math books will need to be rewritten. :-)
>
> I assume the expression would short-circuit once one of the
> sub-expressions returned false, since
>
> x < y < z
>
> cannot be true unless x < y. So there would never be a false < z
> comparison.

So...

class Numeric
alias :old_lt :<
alias :old_gt :<

def <(val)
val if self.old_lt val
false
end

def >(val)
val if self.old_gt val
false
end
end

>> x = 5
>> puts 4 < x < 6
=> 5
>> puts 6 < x < 7
=> false

?

> David

E