In Haskell, is there a way to restrict a monad M a
so that a
satisfy a type class constraint?
I am translating the probabilistic modeling example from F# to Haskell. However, in Haskell, I omitted support
because it would change data Distribution a
to data (Ord a) => Distribution a
. With this change, I get the following error:
...probabilisticModeling.hs:42:13:
Could not deduce (Ord a) from the context ()
arising from a use of `always'
at ...probabilisticModeling.hs:42:13-18
Possible fix:
add (Ord a) to the context of the type signature for `return'
In the expression: always
In the definition of `return': return = always
In the instance declaration for `Monad Distribution'
Indeed, the type of always
/return
is: (Ord a) => a -> Distribution a
. Is there a way I can have a monad Distribution
, but force the constraint (Ord a)
on this monad? I tried:
instance Monad Distribution where
(>>=) = bind
return :: (Ord a) => a -> Distribution a = always
But I get the error:
...probabilisticModeling2.hs:48:4:
Pattern bindings (except simple variables) not allowed in instance declarations
return :: (Ord a) => a -> Distribution a = always
Failed, modules loaded: none.
So it there a way to have a monad M a
, but restrict the a
with a constraint such as Ord a
?
Thanks.
The Monad type class The return function takes a value and returns a monadic value, i.e. a value wrapped in the monad type constructor. The >>= operator (pronounced “bind”) is a crucial part of a monad. It binds a value returned from a computation to another computation.
Lists are a fundamental part of Haskell, and we've used them extensively before getting to this chapter. The novel insight is that the list type is a monad too! As monads, lists are used to model nondeterministic computations which may return an arbitrary number of results.
Since this isn't possible in F#, every time you want to create a free monad for a functor type, you'll have to write the whole thing out with at least one hardcoded reference to the functor type. All of these have (or can have) the same type, Nest<int> .
It appears that I ran into a well-known problem in Haskell. I found many workarounds by googling for "restricted monads". This solutions seems to be the least disruptive. Still, for my purposes, it seems overkill. I think I'll keep the Distribution
monad general, and simplify a support via a restricted function, as suggested by Revolucent.
My understanding of this is that you simply cannot, because a monad is meant to be generalized over all types, not some restricted subset of types such as (Ord a)
.
Instead of restricting the monadic type M a
, you can simply restrict functions which use that monadic type, e.g.,
foo :: Ord a => Int -> M a
In fact, it is preferable to keep types as general as possible and use type classes only to restrict functions.
etc.
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