deriving

Are all differentiable types Monads

由 ￡可爱￡侵袭症＋ 提交于 2019-12-04 08:59:22
问题 Given a differentiable type, we know that its Zipper is a Comonad. In response to this, Dan Burton asked, "If derivation makes a comonad, does that mean that integration makes a monad? Or is that nonsense?". I'd like to give this question a specific meaning. If a type is differentiable, is it necessarily a monad? One formulation of the question would be to ask, given the following definitions data Zipper t a = Zipper { diff :: D t a, here :: a } deriving instance Diff t => Functor (Zipper t)

How to derive instances for records with type-families

由 折月煮酒 提交于 2019-11-30 22:23:01
Here's what I'm trying but it doesn't compile: {-# LANGUAGE TypeFamilies #-} {-# LANGUAGE StandaloneDeriving #-} {-# LANGUAGE FlexibleInstances #-} import Data.Text as T import Data.Int (Int64) type family Incoming validationResult baseType type instance Incoming Validated baseType = baseType type instance Incoming ValidationErrors baseType = Either [T.Text] baseType data Validated data ValidationErrors data Tag = Tag {unTag :: T.Text} deriving (Eq, Show) data NewTag f = NewTag { ntClientId :: Incoming f Int64 , ntTag :: Incoming f Tag } deriving instance (Show baseType) => Show (Incoming

How to derive instances for records with type-families

由 ﹥＞﹥吖頭↗ 提交于 2019-11-30 18:14:55
问题 Here's what I'm trying but it doesn't compile: {-# LANGUAGE TypeFamilies #-} {-# LANGUAGE StandaloneDeriving #-} {-# LANGUAGE FlexibleInstances #-} import Data.Text as T import Data.Int (Int64) type family Incoming validationResult baseType type instance Incoming Validated baseType = baseType type instance Incoming ValidationErrors baseType = Either [T.Text] baseType data Validated data ValidationErrors data Tag = Tag {unTag :: T.Text} deriving (Eq, Show) data NewTag f = NewTag { ntClientId :

Deriving from classes generated by Entity Framework in C#

由 血红的双手。 提交于 2019-11-28 08:13:15
问题 I have created an entity data model and generated a database from it. One of the entities is called Template . Created partial classes to extend the functionality of Template works fine. If I create a new class and try to derive from Template , I get a runtime exception upon instantiating: Mapping and metadata information could not be found for EntityType 'Template001' . How can I work around this? I definitely need to inherit from the EF classes. EDIT Does not seem possible. If that is the

How does deriving work in Haskell?

由 ℡╲_俬逩灬. 提交于 2019-11-27 00:04:05
Algebraic Data Types (ADTs) in Haskell can automatically become instances of some typeclasse s (like Show , Eq ) by deriving from them. data Maybe a = Nothing | Just a deriving (Eq, Ord) My question is, how does this deriving work, i.e. how does Haskell know how to implement the functions of the derived typeclass for the deriving ADT? Also, why is deriving restricted to certain typeclasses only? Why can't I write my own typeclass which can be derived? sclv The short answer is, magic :-). This is to say that automatic deriving is baked into the Haskell spec, and every compiler can choose to

Zipper Comonads, Generically

由 独自空忆成欢 提交于 2019-11-26 21:21:37
Given any container type we can form the (element-focused) Zipper and know that this structure is a Comonad. This was recently explored in wonderful detail in another Stack Overflow question for the following type: data Bin a = Branch (Bin a) a (Bin a) | Leaf a deriving Functor with the following zipper data Dir = L | R data Step a = Step a Dir (Bin a) deriving Functor data Zip a = Zip [Step a] (Bin a) deriving Functor instance Comonad Zip where ... It is the case that Zip is a Comonad though the construction of its instance is a little hairy. That said, Zip can be completely mechanically

How does deriving work in Haskell?

由 落爺英雄遲暮 提交于 2019-11-26 09:16:57
问题 Algebraic Data Types (ADTs) in Haskell can automatically become instances of some typeclasse s (like Show , Eq ) by deriving from them. data Maybe a = Nothing | Just a deriving (Eq, Ord) My question is, how does this deriving work, i.e. how does Haskell know how to implement the functions of the derived typeclass for the deriving ADT? Also, why is deriving restricted to certain typeclasses only? Why can\'t I write my own typeclass which can be derived? 回答1: The short answer is, magic :-).

Zipper Comonads, Generically

由 自闭症网瘾萝莉.ら 提交于 2019-11-26 07:55:08
问题 Given any container type we can form the (element-focused) Zipper and know that this structure is a Comonad. This was recently explored in wonderful detail in another Stack Overflow question for the following type: data Bin a = Branch (Bin a) a (Bin a) | Leaf a deriving Functor with the following zipper data Dir = L | R data Step a = Step a Dir (Bin a) deriving Functor data Zip a = Zip [Step a] (Bin a) deriving Functor instance Comonad Zip where ... It is the case that Zip is a Comonad though
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