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up vote 1 down vote favorite

Here's how I implemented concat - flattens a list of lists into just a list of elements.

concat' :: [[a]] -> [a]
concat' ys = foldl (\acc x -> acc ++ x) [] ys

Is it a problem that I'm using the ++ function?

What's a better way to write this?

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1 Answer 1

up vote 2 down vote

You could eta-contract the definition in two places: \acc x -> acc ++ x is the eta-expanded version of (++) and concat' ys = (...) ys is the eta-expanded version of concat' = (...). So that would take you to:

concat' :: [[a]] -> [a]
concat' = foldl (++) []

Now, in that case it turns out that it's better to use foldr rather than foldl and there is a good write-up about concat on the haskell wiki.

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