Haskell：Show的新实例声明 [英] Haskell: New instance declaration for Show

问题描述

我试图在Haskell中添加一个实例声明，以获得我创建的新数据类型失败。在这里，我已经尝试到目前为止：

pre $ data Prediction = Prediction Int Int Int
showPrediction :: Prediction - > ; String
showPrediction（Prediction a b c）= show a ++ - ++ show b ++ - ++ show c
instance Show（Prediction p）=> showPrediction p


看起来最后一行是错误的，但我不知道如何实现我想要的。基本上就是能够从解释器调用Prediction变量并使其可视化而不必调用showPrediction。现在这个工作：

  showPrediction（Prediction 1 2 3）
  

并显示：

 1-2-3
  

如预期的那样，但我希望这可以起作用（来自解释器）：

 预测1 2 3 
  

任何想法？

解决方案

要派生一个实例，语法是

 实例«preconditions»=> Class«type»其中
«method»=«definition»
  

，你会有

  instance Show Prediction where 
 show（Prediction abc）= show a ++ -  ++ show b ++ - ++ show c 
  

没有先决条件;你会用它来做类似于的实例Show a =>显示[a]其中... ，表示 if a 是可显示的，那么并[a] 。在这里，所有预测都是可显示的，所以没有什么可担心的。当你写实例Show（Prediction p）=> showPrediction p ，你犯了一些错误。首先， Prediction p 意味着 Prediction 是一个参数化类型（例如由数据预测a =预测aaa ），事实并非如此。其次， Show（Prediction p）=> 意味着 if Prediction P ，然后你想声明一些其他的实例。第三，在 => 之后，有一个函数是没有意义的 - Haskell想要一个类型类名。

另外，为了完整起见，如果希望 Prediction 1 2 3 格式用于显示输出，还有另一种派生 Show 的方法：

  data Prediction = Prediction Int Int推导显示
  

按照Haskell 98报告，只有少数类型可以通过这种方式获得： Eq ， Ord ， Enum ，有界，显示和读。通过适当的GHC扩展，您也可以派生 Data ， Typeable ， Functor ，可折叠和可穿透;你可以派生任何一个为 newtype 派生的 newtype 的包装类型的类。并且您可以以独立方式生成这些自动实例。

I'm trying to add an instance declaration in Haskell for a new data type I've created unsuccessfully. Here what I've tried so far:

data Prediction = Prediction Int Int Int
showPrediction :: Prediction -> String
showPrediction (Prediction a b c) = show a ++ "-" ++ show b ++ "-" ++ show c
instance Show (Prediction p) => showPrediction p

Seems the last line is wrong but I'm not sure how to achieve what I want. Basically is to be able to call from the interpreter a Prediction variable and get it visualized without having to call the showPrediction. Right now this works:

showPrediction (Prediction 1 2 3)

and shows:

"1-2-3"

as expected, but I would like this to work (from the interpreter):

Prediction 1 2 3

Any ideas?

解决方案

To derive an instance, the syntax is

instance «preconditions» => Class «type» where
  «method» = «definition»

So here, for instance, you'd have

instance Show Prediction where
  show (Prediction a b c) = show a ++ "-" ++ show b ++ "-" ++ show c

There's no precondition; you'd use that for something like instance Show a => Show [a] where ..., which says that if a is showable, then so is [a]. Here, all Predictions are showable, so there's nothing to worry about. When you wrote instance Show (Prediction p) => showPrediction p, you made a few mistakes. First, Prediction p implies that Prediction is a parametrized type (one declared by, for instance, data Prediction a = Prediction a a a), which it isn't. Second, Show (Prediction p) => implies that if Prediction P is showable, then you want to declare some other instance. And third, after the =>, having a function is nonsensical—Haskell wanted a type class name.

Also, for completeness's sake, there's another way to derive Show if you want the Prediction 1 2 3 format for displayed output:

data Prediction = Prediction Int Int Int deriving Show

As specified in the Haskell 98 report, there are only a handful of types which can be derived this way: Eq, Ord, Enum, Bounded, Show, and Read. With the appropriate GHC extensions, you can also derive Data, Typeable, Functor, Foldable, and Traversable; you can derive any class which a newtype's wrapped type derived for a newtype; and you can generate these automatic instances in a standalone way.

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