functor

来源： https://stackoverflow.com/questions/54261967/what-is-representable-used-for-in-haskell

来源： https://stackoverflow.com/questions/61090254/when-are-template-arguments-required-in-c

来源： https://stackoverflow.com/questions/3273373/are-all-haskell-functors-endofunctors

问题 Consider the following code in C#. public int Foo(int a) { // ... } // in some other method int? x = 0; x = Foo(x); The last line will return a compilation error cannot convert from 'int?' to 'int' which is fair enough. However, for example in Haskell there is Maybe which is a counterpart to Nullable in C#. Since Maybe is a Functor I would be able to apply Foo to x using fmap . Does C# have a similar mechanism? 回答1: We can implement such functionality ourselves: public static class FuncUtils

问题 Consider the following code in C#. public int Foo(int a) { // ... } // in some other method int? x = 0; x = Foo(x); The last line will return a compilation error cannot convert from 'int?' to 'int' which is fair enough. However, for example in Haskell there is Maybe which is a counterpart to Nullable in C#. Since Maybe is a Functor I would be able to apply Foo to x using fmap . Does C# have a similar mechanism? 回答1: We can implement such functionality ourselves: public static class FuncUtils

问题 class Applicative f => Monad f where return :: a -> f a (>>=) :: f a -> (a -> f b) -> f b (<*>) can be derived from pure and (>>=) : fs <*> as = fs >>= (\f -> as >>= (\a -> pure (f a))) For the line fs >>= (\f -> as >>= (\a -> pure (f a))) I am confused by the usage of >>= . I think it takes a functor f a and a function, then return another functor f b . But in this expression, I feel lost. 回答1: Lets start with the type we're implementing: (<*>) :: Monad f => f (a -> b) -> f a -> f b (The

问题 Follow up on this question about Learn You a Haskell for Great Good . The author, at the end of Chapter 8 declares this datatype (slighly simplified, I hope it's fine) data Barry t k p = BarryV p (t k) deriving (Show) and then makes it an instance of Functor instance Functor (Barry a b) where fmap f (BarryV x y) = BarryV (f x) y then concluding There we go! We just mapped the f over the first field. Yes. The first. So my question is: what if I want to map over, say, the second field? Actually

问题 Follow up on this question about Learn You a Haskell for Great Good . The author, at the end of Chapter 8 declares this datatype (slighly simplified, I hope it's fine) data Barry t k p = BarryV p (t k) deriving (Show) and then makes it an instance of Functor instance Functor (Barry a b) where fmap f (BarryV x y) = BarryV (f x) y then concluding There we go! We just mapped the f over the first field. Yes. The first. So my question is: what if I want to map over, say, the second field? Actually

问题 Follow up on this question about Learn You a Haskell for Great Good . The author, at the end of Chapter 8 declares this datatype (slighly simplified, I hope it's fine) data Barry t k p = BarryV p (t k) deriving (Show) and then makes it an instance of Functor instance Functor (Barry a b) where fmap f (BarryV x y) = BarryV (f x) y then concluding There we go! We just mapped the f over the first field. Yes. The first. So my question is: what if I want to map over, say, the second field? Actually

问题 The category of sets is both cartesian monoidal and cocartesian monoidal. The types of the canonical isomorphisms witnessing these two monoidal structures are listed below: type x + y = Either x y type x × y = (x, y) data Iso a b = Iso { fwd :: a -> b, bwd :: b -> a } eassoc :: Iso ((x + y) + z) (x + (y + z)) elunit :: Iso (Void + x) x erunit :: Iso (x + Void) x tassoc :: Iso ((x × y) × z) (x × (y × z)) tlunit :: Iso (() × x) x trunit :: Iso (x × ()) x For the purposes of this question I