[Haskell-cafe] Re: Shouldn't this loop indefinitely => take (last
ndmitchell at gmail.com
Fri Apr 4 11:46:22 EDT 2008
> > > 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:
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.
Read this: http://citeseer.ist.psu.edu/45669.html - it argues the
point I am trying to make, but much better.
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