[Haskell-cafe] Re-exports of resourcet in conduit

Antoine Latter aslatter at gmail.com
Sun Jun 3 01:55:07 CEST 2012

Is it possible that you are puuling in two different versions of the
package that defines the MonadThrow class?

That is, package a was built against version 1, but package b was built
against version 2? This would make GHC think the type-class were

This is just a guess - I have not tried what you are trying.
On Jun 2, 2012 6:35 PM, "Myles C. Maxfield" <myles.maxfield at gmail.com>

> To: Michael Snoyman
> CC: haskell-cafe
> Hello,
> I'm having a problem working with the conduit library, and was hoping
> you could help me out.
> Data.Conduit re-exports ResourceT, MonadResource, and MonadThrow (but
> not ExceptionT) from Control.Monad.Trans.Resource. I have a conduit
> which operates on a monad in the MonadThrow class. I am trying to
> figure out which MonadThrow class this should be (the
> Data.Conduit.MonadThrow class, or the
> Montrol.Monad.Trans.Resource.MonadThrow class, since apparently GHC
> doesn't recognize them as the same, even though one is just a
> re-export of the other).
> If a user of this conduit wants to chain this conduit up with
> something like sourceFile, the underlying monad has to be a member of
> Data.Conduit.MonadResource and whatever MonadThrow class I chose to
> use. I would like to be able to use an existing instance to lift the
> class of the inner monad to the class of the entire monad stack (so I
> don't have to tell the user of my conduit that they have to define
> their own instances), and the only rule that I can find that does that
> is the following from Data.Conduit:
> Data.Conduit.MonadThrow m => Data.Conduit.MonadThrow
> (Data.Conduit.ResourceT m)
> However, GHC doesn't seem to think that
> Control.Monad.Trans.Resource.ExceptionT is an instance of
> Data.Conduit.MonadThrow:
>    No instance for (Data.Conduit.MonadThrow (ExceptionT IO))
>      arising from a use of `.....'
> Control.Monad.Trans.Resource has a similar instance:
> Control.Monad.Trans.Resource.MonadThrow m =>
> Control.Monad.Trans.Resource.MonadThrow
> (Control.Monad.Trans.Resource.ResourceT m)
> but because sourceFile operates in the Data.Conduit.MonadResource
> class, and Control.Monad.Trans.Resource.ResourceT isn't a member of
> that class (it's only a member of
> Control.Monad.Trans.Resource.MonadResource), that doesn't help:
>    No instance for (Data.Conduit.MonadResource
>                       (Control.Monad.Trans.Resource.ResourceT (ExceptionT
> IO)))
>      arising from a use of `.....'
> It should be noted that neither module defines anything like the following:
> MonadResource m => MonadResource (ExceptionT m)
> It seems like the underlying problem here is that:
> 1) I am required to use the Control.Monad.Trans.Resource.ExceptionT
> class, because Data.Conduit doesn't re-export it
> 2) I am required to use the Data.Conduit.MonadResource class, because
> sourceFile and others require it
> 3) There doesn't seem to be an existing instance that bridges between the
> two.
> This seems like a fundamental flaw with re-exporting; it can only work
> if you re-export every single last thing from the original module.
> This doesn't seem tenable because the orignal module might not be
> under your control, so its author can add new symbols whenever he/she
> wants to.
> I see two solutions to this problem:
> 1) Re-export Control.Monad.Trans.Resource.ExceptionT in Data.Conduit.
> This will work until someone adds something to the resourcet package
> and someone wants to use the new addition and Data.Conduit.ResourceT
> in the same stack
> 2) Don't re-export anything in Data.Conduit; make sourceFile and
> others explicitly depend on types in another module, but this might
> break compatibility with existing programs if they use fully-qualified
> symbol names.
> 3) Make anyone who wants to use a monad stack in MonadThrow and
> MonadResource define their own instances. This is probably no good
> because it means that many different modules will implement the same
> instance in potentially many different ways.
> I feel like option 2) is probably the best solution here. I'm
> perfectly happy to issue a pull request for whichever option you think
> is best, but I don't know which solution you think is best for your
> project. What do you think?
> --Myles
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