On Sat, 2002-06-29 at 15:26, Mark Carroll wrote:
On Sat, 29 Jun 2002, Samuel E. Moelius III wrote: (snip)
Here's another not-exactly-what-you-wanted solution. :) (snip)
Do any of the experimental extensions to Haskell allow a what-he-wanted solution? I couldn't arrange one in H98 without something having an infinitely-recursive type signature. I'm sure it would have been easy in Lisp, and he already gave a Perl equivalent, so I'm wondering if it could be at all sane for Haskell to allow such stuff and if Haskell is somehow keeping us on the straight and narrow by disallowing the exact counter that was originally requested.
The beauty of his request was that it was so simple and seemed to make sense; I went ahead and tried to fulfill it before realising I couldn't do it either.
I could not manage to do this with a simple always-increment-by-one function, but the problem of adding a number n each time was a quite a bit easier - though it still took me a while to escape the infinite recursive type , it seems that you need to indirect through another datatype (here FP). you can't print z or z', but the show defined will allow you to print out a FooPair ----- data FooPair = FP Integer (Integer -> FooPair) instance Show FooPair where show (FP i f) = "FooPair " ++ (show i) ++ "...fun..." incg :: Integer -> Integer -> FooPair incg n = \i -> let j = n+i in (FP j (incg j)) val (FP i _) = i fun (FP _ f) = f x = incg 7 -- the original function y = x 3 -- increment the current value by 3 and return the FP pair zf = fun y -- get the new function zv = val y -- and the value in the pair z' = z 99 -- now get the next value function pair ----- -- jeff putnam -- jefu.jefu@verizon.net -- http://home1.get.net/res0tm0p