为列表实现 sequenceAL

Question

Typeclassopedia（Haskell Wiki）在 Applicative 部分有这个练习：

Implement a function
sequenceAL :: Applicative f => [f a] -> f [a]

我只能这样做，如果使用 do 表示法（需要将 Monad f 添加到上下文中：

sequenceAL :: (Monad f, Applicative f) => [f a] -> f [a]
sequenceAL [] = pure []
sequenceAL (x:xs) = do
                     y <- x
                     ys <- sequenceAL xs
                     pure ([y] ++ ys)

如何在不使用 do-notation 和 monad 的情况下实现这一目标。

Answer 1

可以在Control.Applicative中使用<$>和<*>运算符来实现:

sequenceAL :: Applicative f => [f a] -> f [a]
sequenceAL [] = pure []
sequenceAL (x:xs) = (:) <$> x <*> sequenceAL xs

为列表实现 sequenceAL

Implementing sequenceAL for lists

haskell

applicative