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How about this: myFoldr :: (a -> b -> b) -> b -> [a] -> b myFoldr f z xs = foldl' (\s x v -> s (x `f` v)) id xs $ z Cheers, Ivan
How about this:
myFoldr :: (a -> b -> b) -> b -> [a] -> b myFoldr f z xs = foldl' (\s x v -> s (x `f` v)) id xs $ z
Cheers, Ivan
Great! Now I really can say "Come on! It's fun! I can write foldr with foldl!"
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