Haskell MonadIO 与 IO

Question

两者有什么区别:

recompile :: MonadIO m => Bool -> m Bool
recompile :: Bool -> IO Bool

Answer 1

类型 forall m. MonadIO m => Bool -> m Bool和 Bool -> IO Bool是同构的:

{-# LANGUAGE RankNTypes #-}
import Control.Monad.IO.Class

from :: (forall m. MonadIO m => Bool -> m Bool) -> (Bool -> IO Bool)
from act = act

to :: (Bool -> IO Bool) -> (forall m. MonadIO m => Bool -> m Bool)
to act = liftIO . act

Not all types involving IO are isomorphic to a version that replaces IO with a constrained m , though.

一般使用 MonadIO m而不是 m ~ IO当他们想散的时候 liftIO通过调用代码会很烦人或当有其他 mtl m 上需要 -style 约束;并使用 m ~ IO而不是 MonadIO m在追求 API 的简单性时( MonadIO m 会导致复杂性，因此不可取)或在处理 fork 或异常处理时(由于链接问题中讨论的原因， MonadIO m 是不可能的)。