monads

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 I have the following sample of code: {-# LANGUAGE ScopedTypeVariables #-} main = do putStrLn "Please input a number a: " a :: Int <- readLn print a putStrLn "Please input a number b: " b :: Int <- readLn print b putStrLn ("a+b+b^2:" ++ (show $ a+b+c)) where c = b^2 For some reason I cannot use variable b in a where clause, the error I get is the following: Main3.hs:13:15: error: Variable not in scope: b | 13 | where c = b^2 | ^ Any ideas how to make b available in the where clause? 回答1: Use

问题 I was reading about list monads and encountered: [1,2] >>= \n -> ['a','b'] >>= \ch -> return (n,ch) it produces [(1,'a'),(1,'b'),(2,'a'),(2,'b')] Here's how I understand it: Implicit parentheses are: ([1,2] >>= \n -> ['a','b']) >>= (\ch -> return (n,ch)) ([1,2] >>= \n -> ['a','b']) should give [('a',1),('b',1),('a',2),('b',2)] because instance Monad [] where return x = [x] xs >>= f = concat (map f xs) -- this line fail _ = [] so concat (map f xs) is concat (map (\n -> ['a','b']) [1,2]) which

问题 I was reading about list monads and encountered: [1,2] >>= \n -> ['a','b'] >>= \ch -> return (n,ch) it produces [(1,'a'),(1,'b'),(2,'a'),(2,'b')] Here's how I understand it: Implicit parentheses are: ([1,2] >>= \n -> ['a','b']) >>= (\ch -> return (n,ch)) ([1,2] >>= \n -> ['a','b']) should give [('a',1),('b',1),('a',2),('b',2)] because instance Monad [] where return x = [x] xs >>= f = concat (map f xs) -- this line fail _ = [] so concat (map f xs) is concat (map (\n -> ['a','b']) [1,2]) which

问题 In ghci: λ> :t (pure 1) (pure 1) :: (Applicative f, Num a) => f a λ> show (pure 1) <interactive>:1:1: No instance for (Show (f0 a0)) arising from a use of `show' Possible fix: add an instance declaration for (Show (f0 a0)) In the expression: show (pure 1) In an equation for `it': it = show (pure 1) λ> pure 1 1 Does this mean that ghci execute Applicative and displays the result, just like IO ? Note that pure () and pure (+1) don't print anything. 回答1: You get the same behaviour if you use

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

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

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

问题 I have this polymorphic code (see this question) with generic monads for model and client: import Control.Monad.Writer class Monad m => Model m where act :: Client c => String -> c a -> m a class Monad c => Client c where addServer :: String -> c () scenario1 :: forall c m. (Client c, Model m) => m () scenario1 = do act "Alice" $ addServer @c "https://example.com" and this is the pretty-print interpreter for the Client that explains the actions in the log via Writer monad: type Printer =

问题 I have this polymorphic code (see this question) with generic monads for model and client: import Control.Monad.Writer class Monad m => Model m where act :: Client c => String -> c a -> m a class Monad c => Client c where addServer :: String -> c () scenario1 :: forall c m. (Client c, Model m) => m () scenario1 = do act "Alice" $ addServer @c "https://example.com" and this is the pretty-print interpreter for the Client that explains the actions in the log via Writer monad: type Printer =