如何让MonadRandom成为Functor？

Question

似乎来自random-fu包的MonadRandom不是Functor，因为我得到的错误如下：

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，而不是类型为* -> *的类型，例如Maybe 。 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

但是 ，在这种情况下， MonadRandom的实例已经是MonadRandom函数，所以现在实例都不明确！ 相反，您应该在函数中添加Functor约束：

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