
3 Nov
2010
3 Nov
'10
9:18 p.m.
On Wed, 2010-11-03 at 18:05 +0100, Petr Pudlak wrote:
Hi Günther,
from the semantical point of view, you can replace
let x = e' in e by (\x -> e) e' Both should evaluate to the same thing.
You also need (sometimes) fix function
let xs = 1:xs in xs
and
fix (\xs -> 1:xs)
Regards