4 Apr
2008
4 Apr
'08
12:37 p.m.
Hi
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?
Yes Thanks Neil