考虑以下示例:
{-# language ApplicativeDo #-}
module X where
data Tuple a b = Tuple a b deriving Show
instance Functor (Tuple a) where
fmap f (Tuple x y) = Tuple x (f y)
instance Foldable (Tuple a) where
foldr f z (Tuple _ y) = f y z
instance Traversable (Tuple a) where
traverse f (Tuple x y) = do
y' <- f y
let t' = Tuple x y'
return $ t'
看起来不错!但不是:
[1 of 1] Compiling X ( X.hs, interpreted )
X.hs:15:9: error:
• Could not deduce (Monad f) arising from a do statement
from the context: Applicative f
bound by the type signature for:
traverse :: forall (f :: * -> *) a1 b.
Applicative f =>
(a1 -> f b) -> Tuple a a1 -> f (Tuple a b)
at X.hs:14:5-12
Possible fix:
add (Monad f) to the context of
the type signature for:
traverse :: forall (f :: * -> *) a1 b.
Applicative f =>
(a1 -> f b) -> Tuple a a1 -> f (Tuple a b)
• In a stmt of a 'do' block: y' <- f y
In the expression:
do y' <- f y
let t' = Tuple x y'
return $ t'
In an equation for ‘traverse’:
traverse f (Tuple x y)
= do y' <- f y
let t' = ...
return $ t'
|
15 | y' <- f y
| ^^^^^^^^^
Failed, no modules loaded.
即使失败了:
instance Traversable (Tuple a) where
traverse f (Tuple x y) = do
y' <- f y
let unrelated = 1
return $ Tuple x y'
因此，引入任何
let语句都会从“applicative do”中删除“applicative”。为什么？ 最佳答案
它将转化为
let unrelated = 1 in return $ Tuple x y'
不具有
return <something>形式，而applicative则为requires the last statement to be a return or pure :如果您查看https://gitlab.haskell.org/ghc/ghc/wikis/applicative-do中的desugaring描述，它也不支持
let。关于haskell - 为什么 "let"语句强制 "applicative do"块要求monad约束？，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/58254890/