Before 2.06, the definition of `seq' was
seq x y = y
i.e., not terribly useful :)
--Sigbjorn
Graeme Moss writes:
> (Usual apologies if this has been noted/solved since 2.05.)
>
> The following program (which sums [1..n] by brute force):
>
> >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
> import System(getArgs)
>
> data Test = Test !Int
>
> f :: Test -> Int -> Test
> f (Test n) i = Test (n+i)
>
> g :: Test -> Int
> g (Test n) = n
>
> sfoldl :: Eval a => (a -> b -> a) -> a -> [b] -> a
> sfoldl f z [] = z
> sfoldl f z (x:xs) = sfoldl f fzx (fzx `seq` xs)
> where fzx = f z x
>
> main = do argv <- getArgs
> print (g (sfoldl f (Test 0) [1..(read (argv!!0))]))
> <<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<
>
> blows the heap up when compiled with GHC, but not with HBC or Hugs.
>
> The strict version of foldl will apply seq to (f (Test n) i) which
> should force it to (Test (n+i)) and the (n+i) should be evaluated
> because of the strictness annotation. Thus the program should run in
> constant space (indeed with a heap of just 32Kb, compiled with HBC, it
> can evaluate the sum of [1..50000]).
>
> I'm using GHC 2.05.
>
> Graeme.
