Question

假设我有两个这样的函数：

food :: Eatable a => String -> a
food animalType = getAnimal animalType

getAnimal :: Eatable a => String -> a
getAnimal "cat" = Cat
getAnimal "dog" = Dog

Cat和Dog都是Eatable s。

所以我可以像这样调用食物功能：

let cat = food "cat" :: Cat

但是那里的字符串似乎是多余的。有没有办法将其改为：

let cat = food :: Cat

Answer 1

是的，只需删除多余的参数即可。返回类型getAnimal确定将使用的实例。

data Cat = Cat
data Dog = Dog

class Eatable animal where
  getAnimal :: animal

instance Eatable Dog where
  getAnimal = Dog

instance Eatable Cat where
  getAnimal = Cat

有没有办法在此函数中删除类型的冗余？

1 个答案: