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

[Thomas Hartman]

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 <prstanley at ntlworld.com> 
Sent by: haskell-cafe-bounces at haskell.org
11/14/2007 06:46 PM

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haskell-cafe at haskell.org
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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 
please?
Thanks,
Paul

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