Confusion regarding laziness
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by Abhinav Kaushik
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Published on 2010-05-25T11:14:39Z
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2010/05/25
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Filed under:
haskell
|functional-programming
I have a function
myLength = foldl (\ x _ -> x + 1) 0
which fails with stack overflow with input around 10^6 elements (myLength [1..1000000] fails). I believe that is due to the thunk build up since when I replace foldl with foldl', it works. So far so good.
But now I have another function to reverse a list :
myReverse = foldl (\ acc x -> x : acc) []
which uses the lazy version foldl (instead of foldl')
When I do
myLength . myReverse $ [1.1000000].
This time it works fine. I fail to understand why foldl works for the later case and not for former?
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