我如何使Scala中的lambda函数具有通用性？ [英] How do I make lambda functions generic in Scala?

问题描述

使得'scala'中的' def'类泛型是相当直接的

  def someFunc [T]（a：T）{//在此处插入
  

我在这里遇到的问题是如何进行如下通用操作：

  val someFunc =（a：Int）=> //在此处插入正文
  

当然现在a是一个整数，但是我需要做什么使其具有通用性？

val someFunc [T] =（a：T）=> 不会工作， val someFunc = [T]（a：T）=>

使它们变得通用，或者我应该坚持'def'变体？

正如Randall Schulz所说，<$ c

$ c> def 不会创建函数，而是创建一个方法。但是，它可以返回一个函数，这样就可以在 Predef 中创建通用函数，如 identity 函数。这看起来像这样：

  def myId [A] =（a：A）=> a 
 
列表（1,2,3）map myId 
 //列表（1,2,3）
 
列表（foo）map myId 
 // List（foo）
  

但请注意，调用<$ c $没有任何类型信息的c> myId 推断 Nothing 。在上面的例子中，它是有效的，因为类型推断使用 map 的签名，它是 map [B]（f：A => B ），其中 A 是列表的类型，并且 B 被传递给相同作为 A ，因为这是 myId 的签名。

As most of you probably know you can define functions in 2 ways in scala, there's the 'def' method and the lambda method...

making the 'def' kind generic is fairly straight forward

def someFunc[T](a: T) { // insert body here

what I'm having trouble with here is how to make the following generic:

val someFunc = (a: Int) => // insert body here

of course right now a is an integer, but what would I need to do to make it generic?

val someFunc[T] = (a: T) => doesn't work, neither does val someFunc = [T](a: T) =>

Is it even possible to make them generic, or should I just stick to the 'def' variant?

解决方案

As Randall Schulz said, def does not create a function, but a method. However, it can return a function and this way you can create generic functions like the identity function in Predef. This would look like this:

def myId[A] = (a: A) => a

List(1,2,3) map myId
// List(1,2,3)

List("foo") map myId
// List("foo")

But be aware, that calling myId without any type information infers Nothing. In the above case it works, because the type inference uses the signature of map, which is map[B](f: A => B) , where A is the type of the list and B gets infered to the same as A, because that is the signature of myId.

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