On Mon, 2005-04-18 at 01:08 -0400, David Menendez wrote: > Trevion writes: > > > On 4/18/05, Lloyd Allison <[EMAIL PROTECTED]> > wrote: > > > Is it possible to define Y in Haskell (pref' H98) -- > > > and get it to pass type checking? > > > > I believe that > > > > y f = f (y f) > > > > is the normal way to do it. > > I've also seen this: > > y f = g > where g = f g
Note that the second form generally gives more sharing than the first (I say generally because there is no guarantee in Haskell how much sharing you will get.) At least GHC (and probably hugs and nhc98) gives the second form a cyclic representation. See the discussion in Section 4.1 (Recursion) in "A Natural Semantics for Lazy Evaluation", John Launchbury. Bernie. _______________________________________________ Haskell mailing list Haskell@haskell.org http://www.haskell.org/mailman/listinfo/haskell
