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Conduit の join

Haskell

Posted at 2012-02-01

ResourceT m (Source m a) -> Source m a
なる関数が書けてしまったのですが、これは正しいのでしょうか？
一応ちょっと使った感じ動作しているようには見えるのですが…。

import qualified Data.Conduit as C
import qualified Data.Conduit.List as CL

join :: C.ResourceIO m => C.ResourceT m (C.Source m a) -> C.Source m a
join m = C.Source (C.Open <$> n <*> pure []) (return ()) C.$= CL.concatMap id
  where
    n = fmap (\a -> [a]) <$> m

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