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

问题描述

我试图在Haskell中为我创建失败的新数据类型添加实例声明.这是到目前为止我尝试过的:

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

似乎最后一行是错误的，但是我不确定如何实现我想要的.基本上是能够从解释器中调用Prediction变量并使其可视化，而无需调用showPrediction.现在这行得通:

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)

并显示:

"1-2-3"

如预期的那样，但是我希望它能工作(来自解释程序):

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

Prediction 1 2 3

有什么想法吗?

推荐答案

要派生实例，语法为

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

例如，在这里，您将拥有

So here, for instance, you'd have

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

没有先决条件.您可以将其用于类似instance Show a => Show [a] where ...的内容，该内容表示如果 a是可显示的，那么[a]也是如此.在这里，所有Predictions都是可显示的，因此无需担心.编写instance Show (Prediction p) => showPrediction p时，您犯了一些错误.首先，Prediction p表示Prediction是参数化类型(例如，由data Prediction a = Prediction a a a声明的一种)，但不是.其次，Show (Prediction p) =>表示 if Prediction P是可显示的， then 您要声明其他实例.第三，在=>之后，拥有一个函数是荒谬的— Haskell想要一个类型类名称.

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.

另外，出于完整性考虑，如果您想要Prediction 1 2 3格式用于显示的输出，还有另一种派生Show的方法:

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

在Haskell 98报告中指定，只有一个可以通过这种方式派生的少数类型:Eq，Ord，Enum，Bounded，Show和Read.使用相应的GHC扩展名，您还可以导出Data，Typeable，Functor，Foldable和Traversable;您可以派生newtype的包装类型为newtype派生的任何类；您可以以独立方式生成这些自动实例.

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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