[Haskell-cafe] Re: Shouldn't this loop indefinitely => take (last
loup.vaillant at gmail.com
Fri Apr 4 12:31:30 EDT 2008
2008/4/4, Neil Mitchell <ndmitchell at gmail.com>:
> > > > Also, having strict Int's by default is a bit ugly, in an
> > > > otherwise lazy-by-default language.
> > 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?
> length xs == length ys
> 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)?
What do you mean by "proper Lazy naturals"? Peano ones?
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