haskell - 如何使MonadRandom成为仿函数？

似乎来自random-fu软件包的MonadRandom不是Functor，因为我遇到如下错误:

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

但是我得到了错误:

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’

最佳答案

MonadRandom是一个类，而不是类型为* -> *的类型，例如Maybe。通常，您会使用类似

instance MonadRandom m => Functor m where
    fmap = liftM

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

但是，在这种情况下，instances of MonadRandom 已经是仿函数，因此现在的实例是模棱两可的!相反，您应该在函数中添加Functor约束:

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