为什么`newtype` 不适用于这种模式匹配？

静俊

我通过将数据实现为Enum. 而且，我有一个问题只是替换Enum为newtype.

这是我的测试代码。

newtype Fruit = Fruit Int deriving (Eq)

apple = Fruit 0
banana = Fruit 1
grape = Fruit 2
orange = Fruit 3

instance Show Fruit where
  show apple = "Apple"
  show banana = "Banana"
  show grape = "Grape"
  show orange = "Orange"
  show _ = "Fruit"

test =
  map show [apple, banana, grape, orange, (Fruit 5)]

我预计评估test将是["Apple","Banana","Grape","Orange","Fruit"]，但它是["Apple","Apple","Apple","Apple","Apple"]

我认为它变量可能不适用于模式匹配，但我又尝试了一次case，例如：

-- Fails: everything is "Apple"
instance Show Fruit where
  show g = case g of
    apple -> "Apple"
    banana -> "Banana"
    grape -> "Grape"
    orange -> "Orange"
    _ -> "Fruit"

为了确认我的临时文件，我再次尝试使用下一个代码。

-- Fails: Works only for apple/banana. Variable does not work!
instance Show Fruit where
  show (Fruit 0) = "Apple"
  show (Fruit 1) = "Banana"
  show grape = "Grape"
  show orange = "Orange"
  show _ = "Fruit"

我可以得到 ["Apple","Banana","Grape","Grape","Grape"]

好吧，我可以编写代码来实现Enum，但我只想了解为什么这些代码不起作用？

气

这段代码

instance Show Fruit where
  show apple = "Apple"
  show banana = "Banana"
  show grape = "Grape"
  show orange = "Orange"
  show _ = "Fruit"

相当于

instance Show Fruit where
  show x = "Apple"
  show y = "Banana"
  show z = "Grape"
  show w = "Orange"
  show _ = "Fruit"

因为它声明了命名的新变量apple, ...，这些变量会影响全局定义的变量，而与这些变量没有任何关系。

试试吧

instance Show Fruit where
  show x | x == apple = "Apple"
         | x == banana = "Banana"
         | x == grape = "Grape"
         | x == orange = "Orange"
         | otherwise = "Fruit"

要记住的经验法则是：在模式中，所有变量都是模式定义的局部变量。

我建议打开警告-Wall，因为这会指出局部变量对全局变量的影响。

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