Re: Infinite types

Jeffrey A. Scofield wrote:

Say I have the following function, adapted from Pierce, Types and Programming languages:

f n () = (n, f (n + 1))

But a simple modification seems to cures the problem:

newtype W = W (Int, () -> W)

f n () = W (n, f (n + 1))

w2list (W (n,f)) = n:(w2list $ f ())

*Main> take 5 $ w2list (f 7 ()) [7,8,9,10,11] It works in Haskell98. Actually, at first I thought your problem is more complex. I thought it requires existential quantification to get the infinite type:

{-# OPTIONS -fglasgow-exts #-}

-- Emulating Peano integers

data W = forall a. (Show a) => W a

instance Show W where show (W a) = show a

f 0 = W ([]::[()]) f n = case f (n-1) of (W x) -> W [x]

*Main> f 5 [[[[[[]]]]]]