Data.Vector.Mutable似乎需要PrimMonadin ST s和IOmonads 的实例.
类型类定义为 -
-- | Class of primitive state-transformer monads
class Monad m => PrimMonad m where
-- | State token type
type PrimState m
-- | Execute a primitive operation
primitive :: (State# (PrimState m) -> (# State# (PrimState m), a #)) -> m a
-- | Expose the internal structure of the monad
internal :: m a -> State# (PrimState m) -> (# State# (PrimState m), a #)
Run Code Online (Sandbox Code Playgroud)
它们是这样实现的 -
instance PrimMonad IO where
type PrimState IO = RealWorld
primitive = IO
internal (IO p) = p
instance PrimMonad (ST s) where
type PrimState (ST s) = s
primitive = ST
internal (ST p) = p
Run Code Online (Sandbox Code Playgroud)
我根本不了解类型类的任何功能应该做什么,或者实现如何工作.
但是我需要为STT实现它(http://hackage.haskell.org/package/STMonadTrans-0.3.1给出的那个)
STT 有构造函数 STT s m a
在我天真的尝试我试图取代一切ST s有STT s m:
instance Monad m => PrimMonad (STT s m) where
type PrimState (STT s m) = s
primitive = STT
internal (STT p m) = p
Run Code Online (Sandbox Code Playgroud)
但我得到这个错误:
Not in scope: data constructor `STT'
Run Code Online (Sandbox Code Playgroud)
为的定义primitive和internal,尽管使用STT整个程序中已经多次(虽然我想作为一个类型构造?).
我该如何正确实现这个类型类?
(我最终会用这个STT s (Rand g) a)
编辑:我导入Control.Monad.ST.Trans.Internal以获取STT作为数据构造函数,这些是新的错误:(更改internal (STT s m)后internal (STT s))
Couldn't match kind `*' against `ArgKind'
Kind incompatibility when matching types:
m0 :: * -> *
(#,#) (ghc-prim:GHC.Prim.State# (PrimState (STT s m))) :: ArgKind
-> (#)
In the expression: STT
In an equation for `primitive': primitive = STT
Couldn't match type `m'
with `(#,#) (ghc-prim:GHC.Prim.State# (PrimState (STT s m)))'
`m' is a rigid type variable bound by
the instance declaration at src/pimc/PIMC.hs:41:16
Expected type: ghc-prim:GHC.Prim.State# (PrimState (STT s m))
-> (# ghc-prim:GHC.Prim.State# (PrimState (STT s m)), a #)
Actual type: ghc-prim:GHC.Prim.State# s -> m (STTRet s a)
In the expression: p
In an equation for `internal': internal (STT p) = p
Couldn't match type `a' with `STTRet s a'
`a' is a rigid type variable bound by
the type signature for
internal :: STT s m a
-> ghc-prim:GHC.Prim.State# (PrimState (STT s m))
-> (# ghc-prim:GHC.Prim.State# (PrimState (STT s m)), a #)
at src/pimc/PIMC.hs:44:3
Expected type: ghc-prim:GHC.Prim.State# (PrimState (STT s m))
-> (# ghc-prim:GHC.Prim.State# (PrimState (STT s m)), a #)
Actual type: ghc-prim:GHC.Prim.State# s -> m (STTRet s a)
In the expression: p
In an equation for `internal': internal (STT p) = p
Run Code Online (Sandbox Code Playgroud)
原则上,您可能能够为原语编写实现,但不能为内部编写实现,因为它为您的 monad 强加了某种结构,而该结构只能由 IO 和 ST monad 的内部实现来满足。
然而,我的印象是问题出在 Data.Vector.Mutable 模块上,它的要求过于严格。例如,要在 monad m 中使用 IO Vector,原则上只需要将 IO 嵌入到 m 中(即原始方法),反之亦然(即内部方法)。如果这是正确的,他们应该尝试将 PrimMonad 类细分为嵌入部分和同构部分,例如如下:
-- | Class of primitive state-transformer monads
class Monad m => PrimMonadEmbed m where
-- | State token type
type PrimState m
-- | Execute a primitive operation
primitive :: (State# (PrimState m) -> (# State# (PrimState m), a #)) -> m a
class PrimMonadEmbed m => PrimMonad m where
-- | Expose the internal structure of the monad
internal :: m a -> State# (PrimState m) -> (# State# (PrimState m), a #)
Run Code Online (Sandbox Code Playgroud)
就向后兼容性而言,这可能没问题,因为用户代码中不能存在自定义实例。降低要求也将使他们的代码可用于 IO monad 的转换版本,如 StateT Int IO 等。
您可以尝试两件事:
顺便说一句,以下原语实现应该可以工作(比较 STT 的源代码以了解代码中出了什么问题):
{-# LANGUAGE TypeFamilies #-}
{-# LANGUAGE UnboxedTuples #-}
module Test where
import Control.Monad.Primitive
import Control.Monad.ST.Trans
import Control.Monad.ST.Trans.Internal
instance Monad m => PrimMonad (STT s m) where
type PrimState (STT s m) = s
primitive f = STT (\s -> case (f s) of (# s', v #) -> return (STTRet s' v))
internal _ = error "no implementation of internal for STT"
Run Code Online (Sandbox Code Playgroud)