在Scala中，您可以使匿名函数具有默认参数吗? [英] In Scala, can you make an anonymous function have a default argument?

问题描述

这有效:

scala>  def test(name: String = "joe"): Boolean = true 
test: (name: String)Boolean

我希望它可以相同的方式工作:

I expected this to work in the same way:

scala>  val test: String => Boolean = { (name: String = "joe") => true }

console>:1: error: ')' expected but '=' found.

推荐答案

无聊，正确的答案是不可以"，你不能，但是实际上你可以，通过实验性的

The boring, correct answer is no, you can't, but actually you kind of can, with the experimental single abstract method (SAM) synthesis in 2.11.

首先，您需要使用apply方法参数的默认值定义自己的SAM类型:

First you need to define your own SAM type with the default value for the apply method's parameter:

trait Func {
  def apply(name: String = "joe"): Boolean
}

现在，您可以使用函数文字表示法来定义Func(请注意，您需要使用-Xexperimental启动REPL，此步骤才能起作用):

Now you can use the function literal notation to define a Func (note that you'll need to have started the REPL with -Xexperimental for this step to work):

val f: Func = { (name: String) => name == "joe" }

或者只是:

val f: Func = _ == "joe"

然后是通常的东西:

scala> f("joe")
res0: Boolean = true

scala> f("eoj")
res1: Boolean = false

妙语:

scala> f()
res2: Boolean = true

这不完全是您所要求的语法，并且也没有保证它会离开实验状态，即使这样，默认参数也可能不受支持，但我仍然认为它很整洁现在.

It's not exactly the syntax you're asking for, and there are no promises that this will ever leave experimental status, and even if it does, default arguments may not be supported—but I still think it's pretty neat that it works now.

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