rather than ask the role of $! I found it helpful to first grasp the role
of seq, since $! is defined in terms of seq and seq is a "primitive"
operation (no prelude definition, like with IO, it's a "given").
What helped me grasp seq was its role in a strict fold.
Basically, try to sum all the numbers from 1 to a million. Prelude "sum"
probably gives stack overflow (if not, up it to a billion ;) ), and so
will a naive fold, as is explained at
http://www.haskell.org/haskellwiki/Stack_overflow
The code below basically restates what was already on the wiki, but I
found my definitions of foldl' (using seq, bang patterns, and $!) easier
to understand than the definition on the wiki page, and the definition
from Data.List. (Maybe I'll edit the wiki.)
t.
{-# LANGUAGE BangPatterns #-}
-- stack overflow
t1 = myfoldl (+) 0 [1..10^6]
-- works, as do myfoldl'' and myfoldl'''
t2 = myfoldl' (+) 0 [1..10^6]
-- (myfoldl f q ) is a curried function that takes a list
-- If I understand currectly, in this "lazy" fold, this curried function
isn't applied immediately, because
-- by default the value of q is still a thunk
myfoldl f z [] = z
myfoldl f z (x:xs) = ( myfoldl f q ) xs
where q = z `f` x
-- here, because of the definition of seq, the curried function (myfoldl'
f q) is applied immediately
-- because the value of q is known already, so (myfoldl' f q ) is WHNF
myfoldl' f z [] = z
myfoldl' f z (x:xs) = seq q ( myfoldl' f q ) xs
where q = z `f` x
--same as myfoldl'
myfoldl'' f z [] = z
myfoldl'' f !z (x:xs) = ( myfoldl'' f q ) xs
where q = z `f` x
myfoldl''' f z [] = z
myfoldl''' f z (x:xs) = (myfoldl''' f $! q) xs
where q = z `f` x
PR Stanley