Hi, Is lazy evaluation necessary for a functional language to remain functional?
The reason I ask is that because it violates beta-reduction, and also referential transparency (I think). In haskell, we can transform: g x + f x into: f x + g x as both f and g do not change the parameter x. If g always evaluates to a normal form (in both a lazy and a strict world) g x = x but f is defined thus: f x = (\y -> if x /= 0 then x else y/x) And we apply f to 0 (1/0) then f becomes _|_ therefore: 0 + _|_ /= _|_ + 0 Or, does this just become: _|_ = _|_ ? Or, am I missing something totally obvious? Regards, Chris. _______________________________________________ Haskell mailing list Haskell@haskell.org http://www.haskell.org/mailman/listinfo/haskell
