Haskell 中关于 <$> 和 <*> 的优先级混淆

Question

两个示例均来自 http://learnyouahaskell.com/functors-applicative-functors-and-monoids#applicative-functors、

1). (+) <$> (+3) <*> (*100) $ 5

 (+) <$> (+3) <*> (*100) $ 5, the 5 first got applied to (+3) and
 (*100), resulting in 8 and 500. Then, + gets called with 8 and 500,
 resulting in 508.

从第一个例子来看，似乎 <*> 的优先级高于 <$>。

2). (++) <$> Just "johntra" <*> Just "volta"

 (++) <$> Just "johntra" <*> Just "volta",   resulting in a value
 that's the same as Just ("johntra"++),and now Just ("johntra"++) <*>
 Just "volta" happens, resulting in Just "johntravolta".

从第二个例子来看，似乎 <$> 的优先级高于 <*>。

那么它们的优先级相同吗？谁能给我一些 explanations/references?

Answer 1

<$> 和 <*> 具有相同的优先级和左结合性。 $ 的最低优先级为零。您可以使用 ghci 浏览有关它们的信息：

λ> :i (<$>)
(<$>) :: Functor f => (a -> b) -> f a -> f b
        -- Defined in ‘Data.Functor’
infixl 4 <$>

λ> :i (<*>)
class Functor f => Applicative (f :: * -> *) where
  ...
  (<*>) :: f (a -> b) -> f a -> f b
  ...
        -- Defined in ‘Control.Applicative’
infixl 4 <*>

现在您可以算出类型，看看它们是如何进行类型检查的。

Answer 2

实际上它们都具有相同的优先级（infixl 4：(<*>) and (<$>)），您可以从左到右阅读它 -

(+) <$> (+3) <*> (*100) $ 5
= ((+) <$> (+3)) <*> (*100) $ 5
= (\ a b -> (a+3) + b) <*> (\ a -> a*100) $ 5
= (\ a -> (a+3) + (a*100)) $ 5
= 8 + 500 = 508

记住在这种情况下我们有 f <*> g = \x -> f x (g x)

Haskell 中关于 <$> 和 <*> 的优先级混淆

Precedence Confusion about <$> and <*> in Haskell

haskell

functor