[Haskell-cafe] [Haskell] mapM with Traversables

Marc Ziegert Coeus at gmx.de
Thu Sep 29 03:20:25 CEST 2011

Hi Thomas,
this should be on the haskell-cafe or haskell-beginners mailing list. Haskell at ... is mainly for announcements.

You have:
> f :: Monad m =>
>      a -> m b
> Data.Traversable.mapM :: (Monad m, Traversable t) =>
>                          (a -> m b) -> t a -> m (t b)

So, if you define g with
> g a = do Data.Traversable.mapM f a
 or in short
> g = Data.Traversable.mapM f
, then the type will be
> g :: (Monad m, Traversable t) =>
>      t a -> m (t b)
instead of
> g :: [a] -> m (Seq b)

Try using ghci to find these things out. It helps to get not confused with the types.

Besides the missing "Monad" context, g misses a generic way to convert between different Traversables, which does not exist. You can only convert from any Traversable (imagine a Tree) "toList"; not all Traversables have a "fromList" function.
For conversion, you might want to use Foldable and Monoid, fold to untangle and mappend to recombine; but any specific "fromList" function will surely  be more efficient.

- Marc

-------- Original-Nachricht --------
> Datum: Wed, 28 Sep 2011 17:27:58 -0600
> Von: thomas burt <thedwards at gmail.com>
> An: haskell at haskell.org
> Betreff: [Haskell] mapM with Traversables

> Hi -
> I have a function, "f :: Monad m => a -> m b", as well as a list of a's.
> I'd
> like to produce a sequence (Data.Sequence) of b's, given the a's:
> g :: [a] -> m (Seq b)
> g a = do Data.Traversable.mapM f a   -- type error!
> I see that "Data.Traversable.mapM f a" doesn't work... is this like asking
> the compiler to infer the cons/append operation from the type signature of
> g?
> Do I need to write my own function that explicitly calls the "append"
> functions from Data.Sequence or can I do something else that would work
> for
> any "g :: Traversable t, Traversable u => t a -> m (u b)" given "f :: a ->
> m
> b"?
> Thanks for any comments!
> Thomas

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