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Implementing monad on recursive type


Implementing monad on recursive type

By : helpcentre.us
Date : October 17 2020, 06:10 PM
it should still fix some issue To overcome your specific challenge, you should try pattern matching on the result of f x.
code :
(>>=) (A x mx) f = 
    case f x of
        A y my -> 
        B y -> 
        None -> 
join :: [[a]] -> [a]
join [ [ x, y ], [z], [], [w, q] ] = [x, y, z, w, q]


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Implementing this monad/type in Haskell?

Implementing this monad/type in Haskell?


By : chad1211
Date : March 29 2020, 07:55 AM
hope this fix your issue I really cannot figure out the syntax necessary for this, and it probably comes from my lack of understanding of how types work. , In terms of syntax, it's
code :
instance Monad (DataPoint x) where
    -- etc
instance Functor (DataPoint x) where
    -- etc
The first argument of `Monad' should have kind `* -> *',
but `DataPoint x dataval' has kind `*'
data DataPoint a = DataPoint {x :: Double,
                              dataval :: a}

instance Functor DataPoint where
    fmap f dataPoint = DataPoint {x = x dataPoint,
                                  dataval = f (dataval dataPoint)}
Implementing an interface for a plain old recursive data type

Implementing an interface for a plain old recursive data type


By : Sunil Parmar
Date : March 29 2020, 07:55 AM
I wish this helpful for you You need to wrap S n patterns in parenthesis. After doing that, your will get compiler errors because Nat is already defined in Prelude. So to compile your code just replace Nat with Natural (or anything else). Though, Z and S constructors are also defined in Prelude so you either need to rename everything to be able to test in REPL easily or use %hide directive.
But at least this code compiles:
code :
module Test

data Natural = Z | S Natural

Eq Natural where
  Z == Z = True
  (S n1) == (S n2) = n1 == n2
  _ == _ = False
Implementing the Monad typeclass with a multi-parameter type

Implementing the Monad typeclass with a multi-parameter type


By : user7399002
Date : March 29 2020, 07:55 AM
it helps some times You should use type lambda
code :
implicit def extractInterface[E] = new Monad[({ type λ[A] = Extract[E, A] })#λ] {
  ...
}
implicit def extractInterface[E] = new Monad[Extract[E, ?]] {
  ...
}
error message when implementing `((->) r)` monad (type mismatch)

error message when implementing `((->) r)` monad (type mismatch)


By : Shivu
Date : March 29 2020, 07:55 AM
hope this fix your issue My fatal misunderstanding was that I thought the error message meant that my implementation had an incorrect type signature. (I thought that because there was a forall in the type signature.) It turns out that's not the actual error, and my error in the implementation was that I needed to apply another r term to both f and m in my implementation. to @luqui and @melpomene for pointing it out to me
Implementing nested/recursive datatypes in the presence of type operators

Implementing nested/recursive datatypes in the presence of type operators


By : user3481330
Date : March 29 2020, 07:55 AM
hope this fix your issue I've been working on my implementation of System F-omega in Rust following Pierce's Types and Programming Languages, and I'm looking for guidance on adding recursive types to my implementation using iso-recursive fold/unfold operators. I have already added in sum, product, record, existential, and universal types.
code :
List = μ ΛX. ΛA. Cons A (X A) | Nil
-- μF unfolds to F(μF)
= (ΛX. ΛA. Cons A (X A) | Nil) (μ ΛX. ΛA. Cons A (X A) | Nil)
-- substitute *correctly*
= ΛA. Cons A ((μ ΛX. ΛA. Cons A (X A) | Nil) A) | Nil
--                                           ^ you dropped this!
-- folding back
List = ΛA. Cons A (List A) | Nil
data Binary a = Leaf a | Branch (Binary (a, a))
type Binary = μ ΛX. ΛA. Leaf A | Branch (X (A, A))
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