How high can you go with fibac before getting a stack overflow? I'm guessing that it's no more than 200000, and that the suspensions have the following structure.

100 000 works fine 110 000 stack overflow

fibac (as you described):

c = (+) a b d = (+) b c e = (+) c d

sfibac:

c = ($!) ((+) a) b d = ($!) ((+) b) c e = ($!) ((+) c) d

i.e. the ($!) is in the wrong place, and is only making the suspension twice as deep. Addition is already strict in both its parameters.

correct, dumb mistake, thanks :)

...

sfibac' n (a,b) | n == 0 = (a,b) | otherwise = sfibac' (n-1) ((,) b $! a+b)

Yeap, got it. Well, in fact my first aproach was not using pairs:

sfibac2 :: IntPos -> IntPos -> IntPos -> IntPos sfibac2 n a b | n == 0 = a | otherwise = sfibac2 (n-1) b $! (a+b)

With pairs I thought I'd write

dumbfibac :: IntPos -> IntPos -> IntPos -> IntPos dumbfibac n a b | n == 0 = a | otherwise = dumbfibac (n-1) $! (b,(a+b))

But this would obviously not work because ($!) reduces its 2nd argument to 'head normal form', not 'normal form'. Anyway IMO it looks way much better then yours sfibac' function :-) So my question is, is there anyway to force an argument to be reduced to *normal form*? a strict fold function defined like,

sfoldl f a [] = a sfoldl f a (x:xs) = (sfoldl f $! (f a x)) xs

would have the same problem as my dumbfibac function. How do you correctly define a strict fold function? J.A.