'ExceptT ResourceT' vs 'ResourceT ExceptT' - haskell

Real World Haskell states that "Transformer stacking order is important". However, I can't seem to figure out if there's a difference between ExceptT (ResourceT m) a and ResourceT (ExceptT m) a. Will they interfere with each other?

In this example, there is no real difference between both orders. The reason being: unlike many transformers including ExceptT, the resource transformer does not “inject” its own doings into the base monad you apply it to, but rather start off the entire action with passing in the release references.
If you write out the types (I'll refer to MaybeT instead of ExceptT for the sake of simplicity; they're obviously equivalent for the purpose of this question) then you have basically
type MaybeResourceT m a = MaybeT (IORef RelMap -> m a)
= IORef RelMap -> m (Maybe a)
type ResourceMaybeT m a = ResourceT (m (Maybe a))
= IORef RelMap -> m (Maybe a)
i.e. actually equivalent types. I suppose you could also show that for the operations.

Related

What's the difference between view and use in lens?

What's the difference between
view :: MonadReader s m => Getting a s a -> m a
and
use :: MonadState s m => Getting a s a -> m a
in Control.Lens.Getter?
Taking a look at the type signatures, view takes a MonadReader (such as ReaderT) and use takes a MonadState (such as StateT). Now, view and use both have the same objective: extract a reasonable value from the thing we're looking at.
MonadReader represents read-only state. We can access the value within using ask. MonadState represents read-write state, which can be retrieved with get. So both view and use query the internal state of the monad given, but view calls ask whereas use calls get. Generally speaking, only one will be applicable to your situation.
A look at the source code for the two functions is not particularly enlightening unless you already understand how lenses are implemented (and if you do, then you probably understand the difference between view and use), so this is a good example of a situation where the type signatures can be much more enlightening than the code itself.
A lens getter gives us a function from its source to its target:
(^.) :: s -> Getting a s a -> a
flip (^.) :: Getting a s a -> s -> a
Any function can be made into a MonadReader computation, with the argument type of the function as the environment type:
asks :: MonadReader s m => (s -> a) -> m a
That being so, (^.) can be generalised to any MonadReader through asks, giving rise to view:
view :: MonadReader s m => Getting a s a -> m a
view g = asks (\s -> s ^. g)
(The definitions I'm using here aren't literally the ones you will find in the Control.Lens.Getter source, but they are equivalent to them as far as results go.)
In a similar way, any function can be made into a MonadState computation that leaves the state unchanged, with the argument type of the function as the state type:
gets :: MonadState s m => (s -> a) -> m a
Accordingly, (^.) can also be generalised to any MonadState through gets, resulting in use:
use :: MonadReader s m => Getting a s a -> m a
use g = gets (\s -> s ^. g)
From another angle, view and use can be seen as variants of asks and gets, respectively, that take a getter as argument, rather than a function directly.
On a final note about view, functions themselves are instances of MonadReader. That being so, view can be used as a prefix/non-operator version of (^.).

Can IO action in negative position give unexpected results?

There seems to be some undocumented knowledge about the difference between Monad IO and IO. Remarks here and here) hint that IO a can be used in negative position but may have unintended consequences:
Citing Snoyman 1:
However, we know that some control flows (such as exception handling)
are not being used, since they are not compatible with MonadIO.
(Reason: MonadIO requires that the IO be in positive, not negative,
position.) This lets us know, for example, that foo is safe to use in
a continuation-based monad like ContT or Conduit.
And Kmett 2:
I tend to export functions with a MonadIO constraint... whenever it
doesn't have to take an IO-like action in negative position (as an
argument).
When my code does have to take another monadic action as an argument,
then I usually have to stop and think about it.
Is there danger in such functions that programmers should know about?
Does it for example mean that running arbitrary continuation-based action may redefine control flow giving unexpected results in ways that Monad IO based interface are safe from?
Is there danger in such functions that programmers should know about?
There is not danger. Quite the opposite, the point Snoyman and Kmett are making is that Monad IO doesn't let you lift through things with IO in a negative positive.
Suppose you want to generalize putStrLn :: String -> IO (). You can, because the IO is in a positive position:
putStrLn' :: MonadIO m => String -> m ()
putStrLn' str = liftIO (putStrLn str)
Now, suppose you want to generalize handle :: Exception e => (e -> IO a) -> IO a -> IO a. You can't (at least not with just MonadIO):
handle' :: (MonadIO m, Exception e) => (e -> m a) -> m a -> m a
handle' handler act = liftIO (handle (handler . unliftIO) (unliftIO act))
unliftIO :: MonadIO m => m a -> IO a
unliftIO = error "MonadIO isn't powerful enough to make this implementable!"
You need something more. If you're curious about how you'd do that, take a look at the implementation of functions in lifted-base. For instance: handle :: (MonadBaseControl IO m, Exception e) => (e -> m a) -> m a -> m a.

Why are monad transformers different to stacking monads?

In many cases, it isn't clear to me what is to be gained by combining two monads with a transformer rather than using two separate monads. Obviously, using two separate monads is a hassle and can involve do notation inside do notation, but are there cases where it just isn't expressive enough?
One case seems to be StateT on List: combining monads doesn't get you the right type, and if you do obtain the right type via a stack of monads like Bar (where Bar a = (Reader r (List (Writer w (Identity a))), it doesn't do the right thing.
But I'd like a more general and technical understanding of exactly what monad transformers are bringing to the table, when they are and aren't necessary, and why.
To make this question a little more focused:
What is an actual example of a monad with no corresponding transformer (this would help illustrate what transformers can do that just stacking monads can't).
Are StateT and ContT the only transformers that give a type not equivalent to the composition of them with m, for an underlying monad m (regardless of which order they're composed.)
(I'm not interested in particular implementation details as regards different choices of libraries, but rather the general (and probably Haskell independent) question of what monad transformers/morphisms are adding as an alternative to combining effects by stacking a bunch of monadic type constructors.)
(To give a little context, I'm a linguist who's doing a project to enrich Montague grammar - simply typed lambda calculus for composing word meanings into sentences - with a monad transformer stack. It would be really helpful to understand whether transformers are actually doing anything useful for me.)
Thanks,
Reuben
To answer you question about the difference between Writer w (Maybe a) vs MaybeT (Writer w) a, let's start by taking a look at the definitions:
newtype WriterT w m a = WriterT { runWriterT :: m (a, w) }
type Writer w = WriterT w Identity
newtype MaybeT m a = MaybeT { runMaybeT :: m (Maybe a) }
Using ~~ to mean "structurally similar to" we have:
Writer w (Maybe a) == WriterT w Identity (Maybe a)
~~ Identity (Maybe a, w)
~~ (Maybe a, w)
MaybeT (Writer w) a ~~ (Writer w) (Maybe a)
== Writer w (Maybe a)
... same derivation as above ...
~~ (Maybe a, w)
So in a sense you are correct -- structurally both Writer w (Maybe a) and MaybeT (Writer w) a
are the same - both are essentially just a pair of a Maybe value and a w.
The difference is how we treat them as monadic values.
The return and >>= class functions do very different things depending
on which monad they are part of.
Let's consider the pair (Just 3, []::[String]). Using the association
we have derived above here's how that pair would be expressed in both monads:
three_W :: Writer String (Maybe Int)
three_W = return (Just 3)
three_M :: MaybeT (Writer String) Int
three_M = return 3
And here is how we would construct a the pair (Nothing, []):
nutin_W :: Writer String (Maybe Int)
nutin_W = return Nothing
nutin_M :: MaybeT (Writer String) Int
nutin_M = MaybeT (return Nothing) -- could also use mzero
Now consider this function on pairs:
add1 :: (Maybe Int, String) -> (Maybe Int, String)
add1 (Nothing, w) = (Nothing w)
add1 (Just x, w) = (Just (x+1), w)
and let's see how we would implement it in the two different monads:
add1_W :: Writer String (Maybe Int) -> Writer String (Maybe Int)
add1_W e = do x <- e
case x of
Nothing -> return Nothing
Just y -> return (Just (y+1))
add1_M :: MaybeT (Writer String) Int -> MaybeT (Writer String) Int
add1_M e = do x <- e; return (e+1)
-- also could use: fmap (+1) e
In general you'll see that the code in the MaybeT monad is more concise.
Moreover, semantically the two monads are very different...
MaybeT (Writer w) a is a Writer-action which can fail, and the failure is
automatically handled for you. Writer w (Maybe a) is just a Writer
action which returns a Maybe. Nothing special happens if that Maybe value
turns out to be Nothing. This is exemplified in the add1_W function where
we had to perform a case analysis on x.
Another reason to prefer the MaybeT approach is that we can write code
which is generic over any monad stack. For instance, the function:
square x = do tell ("computing the square of " ++ show x)
return (x*x)
can be used unchanged in any monad stack which has a Writer String, e.g.:
WriterT String IO
ReaderT (WriterT String Maybe)
MaybeT (Writer String)
StateT (WriterT String (ReaderT Char IO))
...
But the return value of square does not type check against Writer String (Maybe Int) because square does not return a Maybe.
When you code in Writer String (Maybe Int), you code explicitly reveals
the structure of monad making it less generic. This definition of add1_W:
add1_W e = do x <- e
return $ do
y <- x
return $ y + 1
only works in a two-layer monad stack whereas a function like square
works in a much more general setting.
What is an actual example of a monad with no corresponding transformer (this would help illustrate what transformers can do that just stacking monads can't).
IO and ST are the canonical examples here.
Are StateT and ContT the only transformers that give a type not equivalent to the composition of them with m, for an underlying monad m (regardless of which order they're composed.)
No, ListT m a is not (isomorphic to) [m a]:
newtype ListT m a =
ListT { unListT :: m (Maybe (a, ListT m a)) }
To make this question a little more focused:
What is an actual example of a monad with no corresponding transformer (this would help illustrate what transformers can do that just stacking monads can't).
There are no known examples of a monad that lacks a transformer, as long as the monad is defined explicitly as a pure lambda-calculus term, with no side effects and no external libraries being used. The Haskell monads such as IO and ST are essentially interfaces to an external library defined by low-level code. Those monads cannot be defined by pure lambda-calculus, and their monad transformers probably do not exist.
Even though there are no known explicit examples of monads without transformers, there is also no known general method or algorithm for obtaining a monad transformer for a given monad. If I define some complicated monad, for example like this code in Haskell:
type D a = Either a ((a -> Bool) -> Maybe a)
then it is far from obvious how to define a transformer for the monad D.
This D a may look a contrived and artificial example (and it's also not obvious why it is a monad) but there might be legitimate cases for using that monad, which is a "free pointed monad on the Search monad on Maybe".
To clarify: A "search monad on n" is the type S n q a = (a -> n q) -> n a where n is another monad and q is a fixed type.
A "free pointed monad on M" is the type P a = Either a (M a) where M is another monad.
In any case, I just want to illustrate the point. I don't think it would be easy for anyone to come up with the monad transformer for D and then to prove that it satisfies the laws of monad transformers. There is no known algorithm that takes the code of D and outputs the code of its transformer.
Are StateT and ContT the only transformers that give a type not equivalent to the composition of them with m, for an underlying monad m (regardless of which order they're composed.)
Monad transformers are necessary because stacking two monads is not always a monad. Most "simple" monads, like Reader, Writer, Maybe, etc., stack with other monads in a particular order. But the result of stacking, say, Writer + Reader + Maybe, is a more complicated monad that no longer allows stacking with new monads.
There are several examples of monads that do not stack at all: State, Cont, List, Free monads, the Codensity monad, and a few other, less well known monads, like the "free pointed" monad shown above.
For each of those "non-stacking" monads, one needs to guess the correct monad transformer somehow.
I have studied this question for a while and I have assembled a list of techniques for creating monad transformers, together with full proofs of all laws. There doesn't seem to be any system to creating a monad transformer for a specific monad. I even found a couple of monads that have two inequivalent transformers.
Generally, monad transformers can be classified in 6 different families:
Functor composition in one or another order: EitherT, WriterT, ReaderT and a generalization of Reader to a special class of monads, called "rigid" monads. An example of a "rigid" monad is Q a = (H a) -> a where H is an arbitrary (but fixed) contravariant functor.
The "adjunction recipe": StateT, ContT, CodensityT, SearchT, which gives transformers that are not functorial.
The "recursive recipe": ListT, FreeT
Cartesian product of monads: If M and N are monads then their Cartesian product, type P a = (M a, N a) is also a monad whose transformer is the Cartesian product of transformers.
The free pointed monad: P a = Either a (M a) where M is another monad. For that monad, the transformer's type is m (Either a (MT m a)) where MT is the monad M's transformer.
Monad stacks, that is, monads obtained by applying one or more monad transformers to some other monad. A monad stack's transformer is build via a special recipe that uses all the transformers of the individual monads in the stack.
There may be monads that do not fit into any of these cases, but I have seen no examples so far.
Details and proofs of these constructions of monad transformers are in my draft book here https://github.com/winitzki/sofp

MonadTransControl instance for ProxyFast/ProxyCorrect

Using pipes, I'm trying to write an instance of MonadTransControl for the ProxyFast or ProxyCorrect type. This is what I've got:
instance MonadTransControl (ProxyFast a' a b' b) where
data StT (ProxyFast a' a b' b) a = StProxy { unStProxy :: ProxyFast a' a b' b Identity a}
liftWith = undefined
restoreT = undefined
I have no idea how to write liftWith or restoreT. The instances for the other monad transformers all use a function that "swaps" the monads, for example EitherT e m a -> m (EitherT e Identity a), but I couldn't find any such function in pipes. How does the instance for MonadTransControl for ProxyCorrect / ProxyFast look like? Or is it impossible to write one? (If yes, is it possible in pipes 4.0?)
Thanks for the link, and now I can give a better answer.
No, there is no way to implement this, using either version of pipes. The reason why is that MonadTransControl expects a monad transformer to be built on top of a single layer of the underlying base monad. This is true for all the monad transformers that MonadTransControl currently implements, such as:
ErrorT ~ m (Either e r)
StateT ~ s -> m (r, s)
WriterT ~ m (r, w)
ReaderT ~ i -> m r
ListT ~ m [r] -- This version of ListT is wrong, and the true ListT
-- would not work for `MonadTransControl`
However, a Proxy does not wrap a single layer of the base monad. This is true for both pipes versions where you can nest as many layers of the base monad as you want.
In fact, any monad transformer that nests the base monad multiple times will defy a MonadTransControl instance, such as:
FreeT -- from the `free` package
ListT -- when done "right"
ConduitM -- from the `conduit` package
However, just because pipes does not implement MonadTransControl doesn't mean that all hope is lost. pipes-safe implements many of the operations that one would typically expect from MonadTransControl, such as bracketing resource acquisitions, so if you can elaborate on your specific use case I can tell you more if there is an appropriate pipes-based solution for your problem.

Transformation under Transformers

I'm having a bit of difficulty with monad transformers at the moment. I'm defining a few different non-deterministic relations which make use of transformers. Unfortunately, I'm having trouble understanding how to translate cleanly from one effectful model to another.
Suppose these relations are "foo" and "bar". Suppose that "foo" relates As and Bs to Cs; suppose "bar" relates Bs and Cs to Ds. We will define "bar" in terms of "foo". To make matters more interesting, the computation of these relations will fail in different ways. (Since the bar relation depends on the foo relation, its failure cases are a superset.) I therefore give the following type definitions:
data FooFailure = FooFailure String
data BarFailure = BarSpecificFailure | BarFooFailure FooFailure
type FooM = ListT (EitherT FooFailure (Reader Context))
type BarM = ListT (EitherT BarFailure (Reader Context))
I would then expect to be able to write the relations with the following function signatures:
foo :: A -> B -> FooM C
bar :: B -> C -> BarM D
My problem is that, when writing the definition for "bar", I need to be able to receive errors from the "foo" relation and properly represent them in "bar" space. So I'd be fine with a function of the form
convert :: (e -> e') -> ListT (EitherT e (Reader Context) a
-> ListT (EitherT e' (Reader Context) a
I can even write that little beast by running the ListT, mapping on EitherT, and then reassembling the ListT (because it happens that m [a] can be converted to ListT m a). But this seems... messy.
There's a good reason I can't just run a transformer, do some stuff under it, and generically "put it back"; the transformer I ran might have effects and I can't magically undo them. But is there some way in which I can lift a function just far enough into a transformer stack to do some work for me so I don't have to write the convert function shown above?
I think convert is a good answer, and using Control.Monad.Morph and Control.Monad.Trans.Either it's (almost) really simple to write:
convert :: (Monad m, Functor m, MFunctor t)
=> (e -> e')
-> t (EitherT e m) b -> t (EitherT e' m) b
convert f = hoist (bimapEitherT f id)
the slight problem is that ListT isn't an instance of MFunctor. I think this is the author boycotting ListT because it doesn't follow the monad transformer laws though because it's easy to write a type-checking instance
instance MFunctor ListT where hoist nat (ListT mas) = ListT (nat mas)
Anyway, generally take a look at Control.Monad.Morph for dealing with natural transformations on (parts of) transformer stacks. I'd say that fits the definition of lifting a function "just enough" into a stack.

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