为列表实现sequenceAL

Question

Typeclassopedia（Haskell Wiki）在Applicative的部分中进行了此练习：

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

如果使用do表示法，我只能这样做（要求将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和单子的情况下实现这一目标。

Answer 1

它可以在Control.Applicative中使用<$>和<*>运算符来做到这一点：

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