[Haskell-cafe] Re: Shouldn't this loop indefinitely => take (last
ndmitchell at gmail.com
Fri Apr 4 12:37:53 EDT 2008
> > > I meant:
> > > (\x (y :: Int) -> x + 1) 1 (1/0 :: Int) <=> _|_ ?
> > Division by 0 is still an error. What I mean is:
> Yes, but this particular one need not be performed. Will it be?
Oh, sorry, I misread that. Even with current Haskell's Int's that is
lazy enough to work, and won't crash.
> > Where length xs = 1 and ys = 1000. This takes 1000 steps to tell the
> > Int's aren't equal, since we don't have proper lazy naturals. If we
> > did, it would take 2 steps.
> Err, really? I mean, could we calculate this equality without reducing
> length ys to weak head normal form (and then to plain normal form)?
Not without lazy naturals, or some other way of returning the result
of length lazily.
> What do you mean by "proper Lazy naturals"? Peano ones?
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