foldlとfoldl'の差

foldx.hs

myFoldr :: (a -> b -> b) -> b -> [a] -> b
myFoldr _ a [] = a
myFoldr f a (x:xs) = f x $ myFoldr f a xs

myFoldr1 :: (a -> a -> a) -> [a] -> a
myFoldr1 f (x:xs) = myFoldr f x xs

myFoldl :: (b -> a -> b) -> b -> [a] -> b
myFoldl _ a [] = a
myFoldl f a (x:xs) = myFoldl f (f a x) xs

myFoldl' :: (b -> a -> b) -> b -> [a] -> b
myFoldl' _ a [] = a
myFoldl' f a (x:xs) = seq a' myFoldl' f a' xs
  where a' = f a x

myFoldl1 :: (a -> a -> a) -> [a] -> a
myFoldl1 f (x:xs) = myFoldl f x xs

myFoldl1' :: (a -> a -> a) -> [a] -> a
myFoldl1' f (x:xs) = myFoldl' f x xs

sumTest :: (Num a) => [a] -> a
sumTest = myFoldl1 (+)

-- sumTest [1..10000000]
-- 50000005000000
-- (6.61 secs, 2422963880 bytes)

sumTest' :: (Num a) => [a] -> a
sumTest' = myFoldl1' (+)

-- sumTest' [1..10000000]
-- 50000005000000
-- (2.87 secs, 2003390168 bytes)