MonadPlus instance for ContT

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Ross Paterson

24 Jun 2013 24 Jun '13

1:17 p.m.

Alistair Lynn has proposed the following instance: instance (Monoid r, Monad m) => MonadPlus (ContT r m) where mzero = ContT $ const $ return mempty m `mplus` n = ContT $ \ c -> liftM2 mappend (runContT m c) (runContT n c) but this would also be possible: instance (MonadPlus m) => MonadPlus (ContT r m) where mzero = ContT $ const mzero m `mplus` n = ContT $ \ c -> runContT m c `mplus` runContT n c Is one of them better?

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John Wiegley

24 Jun 24 Jun

1:43 p.m.

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Ross Paterson writes:

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Alistair Lynn has proposed the following instance:

instance (Monoid r, Monad m) => MonadPlus (ContT r m) where mzero = ContT $ const $ return mempty m `mplus` n = ContT $ \ c -> liftM2 mappend (runContT m c) (runContT n c)

but this would also be possible:

instance (MonadPlus m) => MonadPlus (ContT r m) where mzero = ContT $ const mzero m `mplus` n = ContT $ \ c -> runContT m c `mplus` runContT n c

Is one of them better?

I would think that allowing ContT to transform any Monad over any Monoid has more utility than only transforming another MonadPlus. But I have no real world data to back this up, just a hunch. -- John Wiegley FP Complete Haskell tools, training and consulting http://fpcomplete.com johnw on #haskell/irc.freenode.net

Henning Thielemann

1:53 p.m.

On Mon, 24 Jun 2013, John Wiegley wrote:

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Ross Paterson writes:

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Alistair Lynn has proposed the following instance:

instance (Monoid r, Monad m) => MonadPlus (ContT r m) where mzero = ContT $ const $ return mempty m `mplus` n = ContT $ \ c -> liftM2 mappend (runContT m c) (runContT n c)

but this would also be possible:

instance (MonadPlus m) => MonadPlus (ContT r m) where mzero = ContT $ const mzero m `mplus` n = ContT $ \ c -> runContT m c `mplus` runContT n c

Is one of them better?

I would think that allowing ContT to transform any Monad over any Monoid has more utility than only transforming another MonadPlus. But I have no real world data to back this up, just a hunch.

I guess that if someone has a MonadPlus monad, then he wants to continue to use it when the monad stack grows by a ContT. This would give preference to the second instance.

Edward Kmett

2:23 p.m.

I agree with Henning. I've historically used the moral equivalent of the second one far more than the first. I have used both though. For instance, Numeric.Covector in the algebra package is a semantically restricted form of Cont that uses the equivalent of the first method. The second form goes out of its way to preserve more of the existing semantics of the transformer stack and I'd dare say that if such an instance was added, it'd be closer to the expected behavior for more users. -Edward On Mon, Jun 24, 2013 at 1:53 PM, Henning Thielemann < lemming@henning-thielemann.de> wrote:

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On Mon, 24 Jun 2013, John Wiegley wrote:

Ross Paterson writes:

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Alistair Lynn has proposed the following instance:

instance (Monoid r, Monad m) => MonadPlus (ContT r m) where mzero = ContT $ const $ return mempty m `mplus` n = ContT $ \ c -> liftM2 mappend (runContT m c) (runContT n c)

but this would also be possible:

instance (MonadPlus m) => MonadPlus (ContT r m) where mzero = ContT $ const mzero m `mplus` n = ContT $ \ c -> runContT m c `mplus` runContT n c

Is one of them better?

I would think that allowing ContT to transform any Monad over any Monoid has more utility than only transforming another MonadPlus. But I have no real world data to back this up, just a hunch.

I guess that if someone has a MonadPlus monad, then he wants to continue to use it when the monad stack grows by a ContT. This would give preference to the second instance.

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Dan Burton

2:53 p.m.

I also agree with Henning. If anything, the 1st instance could be a ContT instance for Monoid. instance (Monoid r, Monad m) => Monoid (ContT r m a) where mempty = ContT $ const $ return mempty m `mappend` n = ContT $ \ c -> liftM2 mappend (runContT m c) (runContT n c) That way both behaviors are easily accessible. Is there any other sensible ContT instance for Monoid? -- Dan Burton On Mon, Jun 24, 2013 at 11:23 AM, Edward Kmett wrote:

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I agree with Henning.

I've historically used the moral equivalent of the second one far more than the first.

I have used both though. For instance, Numeric.Covector in the algebra package is a semantically restricted form of Cont that uses the equivalent of the first method.

The second form goes out of its way to preserve more of the existing semantics of the transformer stack and I'd dare say that if such an instance was added, it'd be closer to the expected behavior for more users.

-Edward

On Mon, Jun 24, 2013 at 1:53 PM, Henning Thielemann < lemming@henning-thielemann.de> wrote:

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On Mon, 24 Jun 2013, John Wiegley wrote:

Ross Paterson writes:

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>> > Alistair Lynn has proposed the following instance:

instance (Monoid r, Monad m) => MonadPlus (ContT r m) where mzero = ContT $ const $ return mempty m `mplus` n = ContT $ \ c -> liftM2 mappend (runContT m c) (runContT n c)

but this would also be possible:

instance (MonadPlus m) => MonadPlus (ContT r m) where mzero = ContT $ const mzero m `mplus` n = ContT $ \ c -> runContT m c `mplus` runContT n c

Is one of them better?

I would think that allowing ContT to transform any Monad over any Monoid has more utility than only transforming another MonadPlus. But I have no real world data to back this up, just a hunch.

I guess that if someone has a MonadPlus monad, then he wants to continue to use it when the monad stack grows by a ContT. This would give preference to the second instance.

______________________________**_________________ Libraries mailing list Libraries@haskell.org http://www.haskell.org/**mailman/listinfo/libraries http://www.haskell.org/mailman/listinfo/libraries

_______________________________________________ Libraries mailing list Libraries@haskell.org http://www.haskell.org/mailman/listinfo/libraries

Edward Kmett

3:07 p.m.

On Mon, Jun 24, 2013 at 2:53 PM, Dan Burton wrote:

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I also agree with Henning.

If anything, the 1st instance could be a ContT instance for Monoid.

instance (Monoid r, Monad m) => Monoid (ContT r m a) where mempty = ContT $ const $ return mempty m `mappend` n = ContT $ \ c -> liftM2 mappend (runContT m c) (runContT n c)

That way both behaviors are easily accessible. Is there any other sensible ContT instance for Monoid?

Yes. The third construction is the 'universal' lifting that works for every Applicative, which would look like: instance (Monad m, Monoid a) => Monoid (ContT r m a) where mempty = return mempty mappend = liftM2 mappend -Edward

John Wiegley

3:20 p.m.

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Edward Kmett writes:

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The third construction is the 'universal' lifting that works for every Applicative, which would look like:

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instance (Monad m, Monoid a) => Monoid (ContT r m a) where mempty = return mempty mappend = liftM2 mappend

Why not then this? That is, once AMP is out... instance (Applicative m, Monoid a) => Monoid (m a) where mempty = pure mempty mappend = liftA2 mappend -- John Wiegley FP Complete Haskell tools, training and consulting http://fpcomplete.com johnw on #haskell/irc.freenode.net

Edward Kmett

3:26 p.m.

Because there are many monads for which you do not want this construction! It isn't what you mean when you use mappend on [a] for sure! It is also different from the behavior for Monad, and it rules out those and many many other perfectly valid uses by overlap. -Edward On Mon, Jun 24, 2013 at 3:20 PM, John Wiegley wrote:

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Edward Kmett writes:

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The third construction is the 'universal' lifting that works for every Applicative, which would look like:

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instance (Monad m, Monoid a) => Monoid (ContT r m a) where mempty = return mempty mappend = liftM2 mappend

Why not then this? That is, once AMP is out...

instance (Applicative m, Monoid a) => Monoid (m a) where mempty = pure mempty mappend = liftA2 mappend

-- John Wiegley FP Complete Haskell tools, training and consulting http://fpcomplete.com johnw on #haskell/irc.freenode.net

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John Wiegley

3:54 p.m.

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Edward Kmett writes:

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Because there are many monads for which you do not want this construction!

It isn't what you mean when you use mappend on [a] for sure! It is also different from the behavior for Monad, and it rules out those and many many other perfectly valid uses by overlap.

Thanks for the clarification, Edward! Very good points. -- John Wiegley FP Complete Haskell tools, training and consulting http://fpcomplete.com johnw on #haskell/irc.freenode.net

Ross Paterson

4:03 p.m.

On Mon, Jun 24, 2013 at 11:53:06AM -0700, Dan Burton wrote:

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If anything, the 1st instance could be a ContT instance for Monoid.

instance (Monoid r, Monad m) => Monoid (ContT r m a) where mempty = ContT $ const $ return mempty m `mappend` n = ContT $ \ c -> liftM2 mappend (runContT m c) (runContT n c)

That way both behaviors are easily accessible.

There's a school of thought that holds that the Monoid and Alternative (and therefore MonadPlus) instances on any functor should agree: http://thread.gmane.org/gmane.comp.lang.haskell.cafe/94642

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participants (5)

Dan Burton
Edward Kmett
Henning Thielemann
John Wiegley
Ross Paterson