I wrote:
Will (id :: A -> A $!) do the trick?
Ulf Norell wrote:
The problem is not with id, it's with composition. For any f and g we have f . g = \x -> f (g x) So _|_ . g = \x -> _|_ for any g.
OK, so then how about f .! g = ((.) $! f) $! g -Yitz _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
