在函数文字中去掉下划线? [英] Leave off underscore in function literal?

问题描述

scala> val alist = List(1,2,3,4,5)
alist: List[Int] = List(1, 2, 3, 4, 5)

scala> alist filter { 2.< }
res2: List[Int] = List(3, 4, 5)

scala> alist filter { 2 < }
res3: List[Int] = List(3, 4, 5)

scala> alist filter { > 3 }
<console>:1: error: ';' expected but integer literal found.
       alist filter { > 3 }

为什么 { 2.<} 和 {2 <} 工作吗?我想至少我应该写 { 2 <_ } 对吧?

Why would { 2.< } and {2 <} work? I think at least I should write { 2 < _ } right?

不需要参数的方法，您也可以省略点号并使用后缀运算符表示法:

A method that requires no arguments, you can alternatively leave off the dot and use postfix operator notation:

scala> val s = "Hello, world!"
s: java.lang.String = Hello, world!
scala> s toLowerCase
res4: java.lang.String = hello, world!

但是这里，< 方法不是那种不需要参数的方法吧?

But here, < method is not those kinds of methods which requires no arguments right?

你能指出这是什么用法吗?

Can you point me what is this usage?

推荐答案

正在发生的是 Eta 扩展 (6.26.5):

What is happening is an Eta Expansion (6.26.5):

Eta-expansion 将方法类型的表达式转换为等效的函数类型的表达.

Eta-expansion converts an expression of method type to an equivalent expression of function type.

在这种情况下，2 < 是一种方法类型:Int 上的方法 <(之一).但是， filter 需要一个函数类型.在这种情况下，Scala 会自动进行 eta 扩展.

In this case, 2 < is a method type: (one of) the method < on Int. However, filter expects a function type. In such a case, Scala does automatic eta expansion.

请注意，因为 filter 预期的类型是已知的，所以它可以正确推断出正在调用的 2 < 方法.

Note that, because the type expected by filter is known, it can correctly infer what 2 < method is being called.

这篇关于在函数文字中去掉下划线?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！