haskell - 如何使 MonadRandom 成为仿函数？

Question

似乎来自 random-fu 包的 MonadRandom 不是 Functor，因为我收到如下错误:

Could not deduce (Functor m) arising from a use of ‘_1’
from the context (MonadRandom m)

我尝试添加以下代码:

instance Functor MonadRandom where
    fmap = liftM

instance Applicative MonadRandom where
    pure  = return
    (<*>) = ap

但我得到了错误:

The first argument of ‘Functor’ should have kind ‘* -> *’,
  but ‘MonadRandom’ has kind ‘(* -> *) -> Constraint’
In the instance declaration for ‘Functor MonadRandom’

The first argument of ‘Applicative’ should have kind ‘* -> *’,
  but ‘MonadRandom’ has kind ‘(* -> *) -> Constraint’
In the instance declaration for ‘Applicative MonadRandom’

Answer 1

MonadRandom是一个类，而不是一种类型 * -> * ，如 Maybe例如。通常，你会使用类似的东西

instance MonadRandom m => Functor m where
    fmap = liftM

instance MonadRandom m => Applicative m where
    pure  = return
    (<*>) = ap

然而 ，在这种情况下 instances of MonadRandom 已经是仿函数，所以现在实例是模棱两可的!相反，您应该添加 Functor约束你的功能:

yourFunction :: (MonadRandom m, Functor m) => ...
-- instead of yourFunction :: (MonadRandom m) => ...