scala - Int上的匹配表达式并不详尽

Question

我已经开始学习Scala。
我很惊讶下一个代码会编译:

object Hello extends App {
  def isOne(num: Int) = num match {
    case 1 => "hello"
  }
}

例如，您不能在Rust中做类似的事情。
为什么Scala编译器不强制我为 case提供默认值？
我会说这有点不安全。
是否有斯卡拉棉短绒或其他东西？也许有些标志？

Answer 1

由于Scala 2.13.4改进了未密封类型(例如Int)的穷举性检查，因此请尝试使用编译器标志

-Xlint:strict-unsealed-patmat

例如

scala -Xlint:strict-unsealed-patmat -Xfatal-warnings
Welcome to Scala 2.13.5 (OpenJDK 64-Bit Server VM, Java 1.8.0_275).
Type in expressions for evaluation. Or try :help.

scala> def isOne(num: Int) = num match {
     |     case 1 => "hello"
     |   }
                             ^
       warning: match may not be exhaustive.
       It would fail on the following input: (x: Int forSome x not in 1)
error: No warnings can be incurred under -Werror.

一般而言，根据 Pattern Matching Expressions

If the selector of a pattern match is an instance of a sealed class,the compilation of pattern matching can emit warnings which diagnosethat a given set of patterns is not exhaustive, i.e. that there is apossibility of a MatchError being raised at run-time.