On Fri 27 Aug, Simon Peyton-Jones wrote:
> > func n = map (func' n) [1..10]
> > func' x y = nfib x
>
> There isn't a free subexpression to lift out of func.
I had always imagined that in a fully lazy language a function like Mike
Thyers example would get transformed into something like this..
func n = map (func' n) [1..10]
func' = \x -> (\y -> nfib x) -- nfib x is free in \y...???
= \x -> let u = nfib x
in \y -> u
So..
func' n = let u = nfib n
in \y -> u
So (nfib n) should get evaluated only once.
Have I gone wrong somewhere?
Regards
--
Adrian Hey
