Int 上的匹配表达式并不详尽

Question

我开始学习 Scala。

我很惊讶下一个代码编译：

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？

我会说它有点不安全。

是否有任何 Scala linter 或其他东西？也许有些旗帜？

Answer 1

嗯，你可以在结构匹配上稍微处理一下，通过设置 scalac 设置中的“-Xfatal-warnings”选项，这会将此警告和其他警告提升为错误。

Answer 2

因为 Scala 2.13.4 改进了非密封类型的穷举检查，例如 Int 所以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.

一般来说，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 diagnose that a given set of patterns is not exhaustive, i.e. that there is a possibility of a MatchError being raised at run-time.

Int 上的匹配表达式并不详尽

Match expression on Int is not exhaustive

scala

compiler-errors

pattern-matching

non-exhaustive-patterns