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Why map does not force strictness whereas zipWith does?

开发者 https://www.devze.com 2023-03-15 17:06 出处:网络
There are two strict versions of zipWith function: 1) Really strict, elements of lists l1 and l2 get evaluated so their thunks do not eat all stack space (Do开发者_如何转开发n Stewart code)

There are two strict versions of zipWith function:

1) Really strict, elements of lists l1 and l2 get evaluated so their thunks do not eat all stack space (Do开发者_如何转开发n Stewart code)

zipWith' f l1 l2 = [ f e1 e2 | (e1, e2) <- zipWith k l1 l2 ]
            where
                k x y = x `seq` y `seq` (x,y)

2) Not really strict, attempt to force evaluation by other way.

zipWith'' f l1 l2 = [ f e1 e2 | (e1, e2) <- zip (map (\x -> x `seq` x) l1) (map (\x -> x `seq` x) l2) ]

The question is: why the equivalent code from the 2nd example using map does not make the function also strict?


It's a common mistake to use

x `seq` x

Which is exactly equivalent to

x

An excellent explanation is available in Neil Mitchell's post on Bad Strictness.


Instead of tautological map, one can use this function to force the list:

evl []     = []
evl (x:xs) = x `seq` (x:evl xs)
-- Cannot figure out how to do this with fold.

Then the strict zipWith is

zipWith''' f xs ys = zipWith f (evl xs) (evl ys)
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