在元组中返回null时,Scala模式匹配会引发匹配错误

Question

下面是两个片段,我无法理解为什么一个成功执行而另一个抛出运行时异常.

片段1:

val str = "HELP"

val perfectTuple: (String, String) = str match {
    case "NO-HELP" => ("First Help", "Second Help")
    case "OTHER-HELP" => ("I won't Help!", "Even,I won't Help!")
    case "HELP" => (null,"Single Help")
    case _ => throw new NoSuchMethodException

  }

Snippet2:

val str = "HELP"

val (firstPart:String, secondPart:String) = str match {
  case "NO-HELP" => ("First Help", "Second Help")
  case "OTHER-HELP" => ("I won't Help!", "Even,I won't Help!")
  case "HELP" => (null,"Single Help")
  case _ => throw new NoSuchMethodException
}

==========================

两个片段之间的差异很小.一个将返回的元组存储到tuple2类型的值"perfectTuple"中,并且这个元素成功执行.

另一个从元组2中提取值并将它们存储到字符串值中并抛出运行时'scala.matchError'.

这是scala中的错误吗？

我在scala 2.10.5和2.11.7上尝试过这个

提前致谢.

=============

还有一个场景,我可以从模式匹配中为字符串赋值null,这个非常完美:Snippet3:

val str = "HELP"

val assignNullToString: String = str match {

    case "NO-HELP" => "ONE"
    case "OTHER-HELP" => "TWO"
    case "HELP" => null
    case _ => throw new NoSuchMethodException
}          

所以我假设,它不是我为String分配null导致问题,它与Tuple有关吗？和Snippet 2有什么问题,而Snippet 1运行得非常好.

Answer 1

那么第二个例子使用的unapply是case class Tuple2.

现在,我没有看过优化的代码,Tuple2.unapply但我猜想在某些时候它会对元组的值进行类型匹配.

一个人根本无法输入匹配null.

val str: Any = null

str match {
  case _: String => "yay"
  case other => "damn"
}
|-> res1: String = damn

更新

让我们稍微讨论你的第二个例子:

val str = "HELP"

val (firstPart:String, secondPart:String) = str match {
  case "NO-HELP" => ("First Help", "Second Help")
  case "OTHER-HELP" => ("I won't Help!", "Even,I won't Help!")
  case "HELP" => (null,"Single Help")
  case _ => throw new NoSuchMethodException
}

当我们提取比赛时,我们得到:

val tuple: (String, String) = str match {
  case "NO-HELP" => ("First Help", "Second Help")
  case "OTHER-HELP" => ("I won't Help!", "Even,I won't Help!")
  case "HELP" => (null,"Single Help")
  case _ => throw new NoSuchMethodException
}

现在,你把它放到了unapply函数中Tuple2.看看签名:

def unapply[A, B](tuple: Tuple2[_, _]): Option[(A, B)]

因此,传入元组值的类型将被删除!但是,当你说

val (first: String, second: String) = tuple

您正在调用Tuple2.unapply类型参数[String, String],明确要求结果(String, String).

为了能够返回Option[(String, String)],unapply函数必须键入两个值匹配.

可以想象Tuple2伴侣对象看起来像,但事实上更有效和复杂:

object Tuple2 {
  def apply[A, B](_1: A, _2: B): Tuple2[A, B] = new Tuple2(_1, _2)

def unapply[A, B](tuple: Tuple2[_, _]): Option[(A, B)] = {
  val a: Option[A] = tuple._1 match { case a: A => Some(a) }
  val b: Option[B] = tuple._2 match { case b: B => Some(b) }

  a.zip(b).headOption
}

现在这里是抛出MatchError的行:

  val a: Option[A] = tuple._1 match { case a: A => Some(a) }

正如我之前所说,我没有看过优化的代码Tuple2,但我有理由相信这与实际发生的情况非常接近.

你当然可以放松一下你的要求:

val str = "HELP"

val (firstPart, secondPart) = str match {
  case "NO-HELP" => ("First Help", "Second Help")
  case "OTHER-HELP" => ("I won't Help!", "Even,I won't Help!")
  case "HELP" => (null,"Single Help")
  case _ => throw new NoSuchMethodException
}

对大多数情况应该足够好,但可能只会推迟疼痛.

编辑2

我建议始终要注意这样一个事实:当使用提取器绑定val时,会使用语法糖进行模式匹配,因为模式匹配可能总是失败并出现匹配错误.

以下是一些明显的例子,我看到过几次不那么明显的事情.

val foo = ("lorem", 2)
val (lorem: String, bad: String) = foo // fail 

case class Bar(name: String, age: Option[Int])
val bar = Bar("Sam", None)
val Bar(name, Some(age)) = bar // fail