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haskell - 如何测试自定义 StateT 的 Monad 实例?

转载 作者:行者123 更新时间:2023-12-05 08:03:00 24 4
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我正在学习 Monad Transformers,其中一个练习要求为 StateT 实现 Monad 实例。我想使用 validity 测试我的实现是否符合 Monad 法则包,类似于 checkers 包。

问题是,我的 Arbitrary 实例无法编译。我看到了this question ,但它并没有完全满足我的要求,因为测试基本上重复了实现并且不检查法律。还有 this question ,但它没有答案,我已经想出如何测试不涉及函数的 Monad 变形金刚(如 MaybeT)。

{-# LANGUAGE DerivingVia #-}
{-# LANGUAGE InstanceSigs #-}

module Ch11.MonadT (StT (..)) where

import Control.Monad.Trans.State (StateT (..))

newtype StT s m a = StT (s -> m (a, s))
deriving
(Functor, Applicative)
via StateT s m

instance (Monad m) => Monad (StT s m) where
return :: a -> StT s m a
return = pure

(>>=) :: StT s m a -> (a -> StT s m b) -> StT s m b
StT x >>= f = StT $ \s -> do
(k, s') <- x s
let StT y = f k
y s'

(>>) :: StT s m a -> StT s m b -> StT s m b
(>>) = (*>)

我的测试:

{-# LANGUAGE DerivingVia #-}
{-# LANGUAGE GeneralizedNewtypeDeriving #-}
{-# LANGUAGE TypeApplications #-}

module Ch11.MonadTSpec (spec) where

import Ch11.MonadT (StT (..))
import Test.Hspec
import Test.QuickCheck
import Test.Validity.Monad

spec :: Spec
spec = do
monadSpecOnArbitrary @(StTArbit Int [] Int)

-- create wrapper to avoid orphan instance error
newtype StTArbit s m a = StTArbit (StT s m a)
deriving
(Functor, Applicative, Monad)

instance (Arbitrary s, Function s, Arbitrary1 m, Arbitrary a) => Arbitrary (StTArbit s m a) where
arbitrary = do
f <- arbitrary :: Fun s (m (a, s))
StTArbit . StT <$> f

错误:

• Couldn't match type: (a0, s0)
with: s -> m (a, s)
Expected: Gen (s -> m (a, s))
Actual: Gen (a0, s0)
• In the second argument of ‘(<$>)’, namely ‘f’
In a stmt of a 'do' block: StTArbit . StT <$> f

最佳答案

OP 在这里,这就是我最终做的。

-- https://ghc.gitlab.haskell.org/ghc/doc/users_guide/exts/explicit_forall.html
{-# LANGUAGE ExplicitForAll #-}
-- https://ghc.gitlab.haskell.org/ghc/doc/users_guide/exts/type_applications.html
{-# LANGUAGE TypeApplications #-}

module Ch11.MonadTSpec (spec) where

import Ch11.MonadT (StT (..), runStT)
import Data.Function as F
import Test.Hspec
import Test.Hspec.QuickCheck
import Test.QuickCheck

spec :: Spec
spec = do
describe "Monad (StT Int [])" $ do
describe "satisfies Monad laws" $ do
-- the types are in the same order as in `forall`
prop "right identity law" (prop_monadRightId @Int @Int @[])
prop "left identity law" (prop_monadLeftId @Int @Int @Int @[])
prop "associative law" (prop_monadAssoc @Int @Int @Int @Int @[])

{- HLINT ignore -}

{-
the types in `forall` are specified in the order of dependency.
since `m` needs `a` and `s`, those appear before `m` in the list.
-}

-- (x >>= return) == x
prop_monadRightId ::
forall a s m.
(Monad m, Eq (m (a, s)), Show (m (a, s))) =>
s ->
Fun s (m (a, s)) ->
Property
prop_monadRightId s f = ((===) `F.on` go) (m >>= return) m
where
m = StT $ applyFun f
go st = runStT st s

-- (return x >>= f) == (f x)
prop_monadLeftId ::
forall a b s m.
(Monad m, Eq (m (b, s)), Show (m (b, s))) =>
a ->
s ->
Fun (a, s) (m (b, s)) ->
Property
prop_monadLeftId a s f = ((===) `F.on` go) (return a >>= h) m
where
g = applyFun2 f
m = StT $ g a
h = StT . g
go st = runStT st s

-- ((x >>= f) >>= g) == (x >>= (\x' -> f x' >>= g))
prop_monadAssoc ::
forall a b c s m.
(Monad m, Eq (m (b, s)), Show (m (b, s)), Eq (m (c, s)), Show (m (c, s))) =>
s ->
Fun s (m (a, s)) ->
Fun (a, s) (m (b, s)) ->
Fun (b, s) (m (c, s)) ->
Property
prop_monadAssoc s h f g =
((===) `F.on` go)
((m >>= f') >>= g')
(m >>= (\x -> f' x >>= g'))
where
m = StT $ applyFun h
f' = StT . applyFun2 f
g' = StT . applyFun2 g
go st = runStT st s

关于haskell - 如何测试自定义 StateT 的 Monad 实例?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/75245910/

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