# role of seq, \$!, and bangpatterns illuminated with lazy versus strict folds Re: [Haskell-cafe] What is the role of \$!?

Thomas Hartman thomas.hartman at db.com
Mon Dec 10 11:46:15 EST 2007

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

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 <prstanley at ntlworld.com>
11/14/2007 06:46 PM

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Subject
[Haskell-cafe] What is the role of \$!?

Hi
What is the role of \$! ?
As far as I can gather it's something to do with strict application.
Could someone explain what it is meant by the term strict application
Thanks,
Paul

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