`sequenceA`如何工作 [英] How `sequenceA` works

问题描述

我是Haskell的新手，正试图了解它是如何工作的?

I'm new to Haskell and trying to understand how does this work?

sequenceA [(+3),(+2),(+1)] 3

我从定义开始

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

然后展开递归

(:) <$> (+3) <*> $ (:) <$> (+2) <*> $ (:) <$> (+1) <*> pure [] 
(:) <$> (+3) <*> $ (:) <$> (+2) <*> $ (:) <$> (+1) <*> []

但是在这里，我不知道对于((->) r)或[]

But here i don't understand for which applicative functor operator <*> will be called, for ((->) r) or for []

(:) <$> (+1) <*> []

有人可以一步一步解析sequenceA [(+3),(+2),(+1)] 3吗?谢谢.

Can somebody go step by step and parse sequenceA [(+3),(+2),(+1)] 3 step by step? Thanks.

推荐答案

它正在使用instance Applicative ((->) a).

在ghci中尝试一下:

Try this in ghci:

Prelude> :t [(+3),(+2),(+1)]
[(+3),(+2),(+1)] :: Num a => [a -> a]

Prelude> :t sequenceA
sequenceA :: (Applicative f, Traversable t) => t (f a) -> f (t a)

和模式匹配参数类型:t = [], f = (->) a 并且应用约束位于f上.

and pattern match the argument type: t = [], f = (->) a and the Applicative constraint is on f.

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