RE: partition and lifted products

2000-01-24 Thread Simon Peyton-Jones
| Can we have extensional products and functions (or at least the means | to define them) please? Does anyone want to come up with a concrete language proposal? Language issues ~~~ A significant difficulty is that seq is essentially un-implementable for unlifted products (requires

Re: partition and lifted products

2000-01-24 Thread S.M.Kahrs
| Can we have extensional products and functions (or at least the means | to define them) please? Language issues ~~~ A significant difficulty is that seq is essentially un-implementable for unlifted products (requires parallelism). Well, all you need is an evaluation with a

Re: partition and lifted products

2000-01-21 Thread Ross Paterson
On Wed, Jan 19, 2000 at 03:18:34PM -0700, Joe Fasel wrote: *Sigh* And the language named in honor of Haskell Curry for which Currying is not a valid transformation strikes again! Worse, not only are the built-in product and function types lifted, but one can't define the unlifted ones.

Re: partition and lifted products

2000-01-21 Thread Mark Tullsen
Ross Paterson wrote: On Wed, Jan 19, 2000 at 03:18:34PM -0700, Joe Fasel wrote: *Sigh* And the language named in honor of Haskell Curry for which Currying is not a valid transformation strikes again! Worse, not only are the built-in product and function types lifted, but one can't

partition and lifted products

2000-01-19 Thread Joe Fasel
Folks, I claimed that these are different functions: partition1 p xs = (filter p xs, filter (not . p) xs) partition2 p = foldr (\x (ys, zs) - if p x then (x:ys,zs) else (ys,x:zs)) ([],[]) I was correct, but not for the reason I thought. Nota bene: partition1 p