applicative

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 Applicative typeclass represents lax monoidal functors that preserve the cartesian monoidal structure on the category of typed functions. In other words, given the canonical isomorphisms witnessing that (,) forms a monoidal structure: -- Implementations left to the motivated reader assoc_fwd :: ((a, b), c) -> (a, (b, c)) assoc_bwd :: (a, (b, c)) -> ((a, b), c) lunit_fwd :: ((), a) -> a lunit_bwd :: a -> ((), a) runit_fwd :: (a, ()) -> a runit_bwd :: a -> (a, ()) The typeclass and its laws

问题 I'm given the following parsers newtype Parser a = Parser { parse :: String -> Maybe (a,String) } instance Functor Parser where fmap f p = Parser $ \s -> (\(a,c) -> (f a, c)) <$> parse p s instance Applicative Parser where pure a = Parser $ \s -> Just (a,s) f <*> a = Parser $ \s -> case parse f s of Just (g,s') -> parse (fmap g a) s' Nothing -> Nothing instance Alternative Parser where empty = Parser $ \s -> Nothing l <|> r = Parser $ \s -> parse l s <|> parse r s ensure :: (a -> Bool) ->

问题 I'm given the following parsers newtype Parser a = Parser { parse :: String -> Maybe (a,String) } instance Functor Parser where fmap f p = Parser $ \s -> (\(a,c) -> (f a, c)) <$> parse p s instance Applicative Parser where pure a = Parser $ \s -> Just (a,s) f <*> a = Parser $ \s -> case parse f s of Just (g,s') -> parse (fmap g a) s' Nothing -> Nothing instance Alternative Parser where empty = Parser $ \s -> Nothing l <|> r = Parser $ \s -> parse l s <|> parse r s ensure :: (a -> Bool) ->

问题 I'm given the following parsers newtype Parser a = Parser { parse :: String -> Maybe (a,String) } instance Functor Parser where fmap f p = Parser $ \s -> (\(a,c) -> (f a, c)) <$> parse p s instance Applicative Parser where pure a = Parser $ \s -> Just (a,s) f <*> a = Parser $ \s -> case parse f s of Just (g,s') -> parse (fmap g a) s' Nothing -> Nothing instance Alternative Parser where empty = Parser $ \s -> Nothing l <|> r = Parser $ \s -> parse l s <|> parse r s ensure :: (a -> Bool) ->

问题 For example, -- Num a => ([Char], a -> a) <*> ([Char], a) > ("hello ",(*6)) <*> ("world",7) ("hello world",42) -- Num a => [a -> a] <*> [a] > [(*7),(*6)] <*> [6,7] [42,49,36,42] -- Num a => [[Char], a -> a] <*> [[Char], a] > ["hello ",(*6)] <*> ["world",7] <interactive>:17:2: Couldn't match expected type ‘[Char] -> [Char]’ with actual type ‘[Char]’ In the expression: "hello " In the first argument of ‘(<*>)’, namely ‘["hello ", (* 6)]’ In the expression: ["hello ", (* 6)] <*> ["world", 7] For

问题 For example, -- Num a => ([Char], a -> a) <*> ([Char], a) > ("hello ",(*6)) <*> ("world",7) ("hello world",42) -- Num a => [a -> a] <*> [a] > [(*7),(*6)] <*> [6,7] [42,49,36,42] -- Num a => [[Char], a -> a] <*> [[Char], a] > ["hello ",(*6)] <*> ["world",7] <interactive>:17:2: Couldn't match expected type ‘[Char] -> [Char]’ with actual type ‘[Char]’ In the expression: "hello " In the first argument of ‘(<*>)’, namely ‘["hello ", (* 6)]’ In the expression: ["hello ", (* 6)] <*> ["world", 7] For

问题 I have this type, basically a Kleisli arrow: {-# language DeriveFunctor #-} data Plan m i o = Plan (i -> m o) deriving Functor instance (Monad m) => Applicative (Plan m i) where pure x = Plan (\_ -> pure x) Plan f <*> Plan x = Plan (\i -> f i <*> x i) Since it has an Applicative instance, I turn on ApplicativeDo and try to build a value using do-notation: {-# language ApplicativeDo #-} myplan :: Plan IO () () myplan = do pure () pure () It doesn't work: No instance for (Monad (Plan IO ()))

问题 I've been playing around with Applicative instances in order to figure out how they work. However, I honestly don't understand this behavior. If I define my own datatype, then apply pure to it with no other arguments, nothing prints out, but it errors if I try to apply something to the result. ghci> data T = A ghci> pure A ghci> pure A 0 <interactive>:21:1: No instance for (Show T) arising from a use of ‘print’ In a stmt of an interactive GHCi command: print it However, if I make T an

问题 I've been playing around with Applicative instances in order to figure out how they work. However, I honestly don't understand this behavior. If I define my own datatype, then apply pure to it with no other arguments, nothing prints out, but it errors if I try to apply something to the result. ghci> data T = A ghci> pure A ghci> pure A 0 <interactive>:21:1: No instance for (Show T) arising from a use of ‘print’ In a stmt of an interactive GHCi command: print it However, if I make T an