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Type signature of foldFree #119

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turion opened this issue Nov 24, 2015 · 0 comments
Open

Type signature of foldFree #119

turion opened this issue Nov 24, 2015 · 0 comments

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@turion
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turion commented Nov 24, 2015

In master/src/Control/Monad/Free.hs, line 323, it says:

 foldFree :: (Functor m, Monad m) => (forall x . f x -> m x) -> Free f a -> m a

It should probably be Functor f, not Functor m. It still typechecks because fmap isn't actually used in the definition, I think. (It would only be used in the proof of correctness, e.g. the naturality square.)

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