为什么我不能在 Scala 中对 Stream.empty 进行模式匹配? [英] Why can't I pattern match on Stream.empty in Scala?

问题描述

如果我取消注释指示的行，下面的代码将无法编译.编译器抱怨:需要稳定的标识符".

The code below doesn't compile if I uncomment the line indicated. The compiler complains: "stable identifier required".

val Empty = Stream.empty    
val a = Stream.range(0, 5)
a match {
  // case Stream.empty => println("nope") <-- does not work
  case Empty => println("compiles") <-- works
  case _ => println("ok")
}

如果我首先将 Stream.empty 赋值给 Empty 值，它可以工作，但是感觉很奇怪，如果没有这样的基本值，您无法对这样的基本值进行模式匹配黑客.

If I assign Stream.empty to value Empty first, it works, but it feels strange that you can't pattern match on such a fundamental value without such a hack.

我错过了什么吗?

推荐答案

您不能对 Stream.empty 进行模式匹配，因为它是一个 方法(在对象 Stream) 总是返回空流(但编译器不知道).

You can't pattern match on Stream.empty because it is a method (in object Stream) that always returns the empty stream (but the compiler doesn't know that).

不是分配 val empty = Stream.empty，你可以匹配 Stream.Empty，它是一个 Object :

Instead of assigning val empty = Stream.empty, you can match on Stream.Empty, which is an Object :

scala> a match {
           case Stream.Empty => println("done")
           case h #:: tl => println(h)
       }

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