Swift中的泛型类型 [英] Generic typealias in Swift

问题描述

在haskell中，您可以这样做：

  type Parser a = String  - > [（a，String）] 
  

我试图在Swift中做类似的事情。到目前为止，我没有运气写这些代码。

  typealias解析器< A> =字符串 - > [（A，String）] 
 typealias解析器a =字符串 - > [（a，String）] 
 typealias Parser = String  - > [（A，String）] 
  

所以这在swift中根本不可能？如果是，还有另外一种方法来实现这种行为？

更新：现在似乎泛型类型别支持swift 3
https://github.com/apple/swift/blob/master/CHANGELOG.md typealias 目前无法与泛型一起使用。您最好的选择可能是将解析器函数封装在结构中。

  struct Parser< A> {
 let f：String  - > [（A，String）] 
} 
  

然后，您可以使用尾随闭包语法创建一个解析器，例如

  let parser = Parser< Character> {string in return [head（string），tail（string）]} 
  

In haskell you can do this:

type Parser a = String -> [(a, String)]

I tried to make something similar in Swift. So far I wrote these codes with no luck.

typealias Parser<A> = String -> [(A, String)]
typealias Parser a = String -> [(a, String)]
typealias Parser = String -> [(A, String)]

So is this simply impossible in swift? And if it is is there another ways to implement this behavior?

UPDATE: It seems generic typealiases are now supported in swift 3 https://github.com/apple/swift/blob/master/CHANGELOG.md

解决方案

typealias cannot currently be used with generics. Your best option might be to wrap the parser function inside a struct.

 struct Parser<A> {
     let f: String -> [(A, String)]
 }

You can then use the trailing closure syntax when creating a parser, e.g.

let parser = Parser<Character> { string in return [head(string), tail(string)] }

这篇关于Swift中的泛型类型的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！