多态类型的显式类型签名 [英] Explicit type signatures for polymorphic types

问题描述

我在回顾 Haskell：函数式编程的工艺，但是第356页的类型签名已经引发了我循环。

这里有一个简单的例子：

  succeed :: b  - >解析ab 
成功val inp = [（val，inp）] 
  

b - >如果


 成功，解析b > Int  - >诠释
成功a = a 
  

和

 成功Int  - > Int  - >诠释
成功ab = a + b 
  

您必须参与的参数数量类型声明权利？如果您的类型声明只有一个类型变量，那么如何取 val 和 imp > 成功:: b - >解析ab 应该读取，取一个 a 类型的变量并返回一个类型的解析ab ，不接受两个变量...其中 inp 允许？

解决方案 Parse 是一个与 - > 类似的扩展类型。例如：

  type Foo = Int  - > Int 
 
 succeed :: Int  - > Foo 
成功a b = a + b 
  

I'm reviewing Haskell: The Craft of Functional Programming, however the type signatures on page 356 have thrown me for a loop.

Here is a simple one:

succeed :: b -> Parse a b
succeed val inp = [( val, inp )]

How can that be b -> Parse a b, if

succeed Int -> Int
succeed a = a

and,

succeed Int -> Int -> Int
succeed a b = a + b

The amount of arguments your taking has to be in the type declaration right? How can you take val, and imp, if your type declaration only has one type variable: succeed :: b -> Parse a b is supposed to read, take one variable of type a and return a type of Parse a b, not take two variables... where is inp permitted?

解决方案

Because Parse is a type synonym with -> in its expansion. Example:

type Foo = Int -> Int

succeed :: Int -> Foo
succeed a b = a + b

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