scala - `Seq.contains` 在类型不匹配时失败的替代方法是什么？

Question

与此类似 Scala: Why does Seq.contains take an Any argument, instead of an argument of the sequence type?

如果你这样做Seq(1, 2, 3).contains("dasdas")
这将编译并始终返回 false。是否有替代方法会引发类型错误？
这似乎是 contains 的主要问题。总是返回 false并且在代码审查中非常容易错过。

Answer 1

您可以启用 -Xfatal-warnings和 -Ywarn-infer-any使其在类型不匹配时失败。
这也适用于平等检查 == .

这是我在 build.sbt 中使用的示例避免像您遇到的奇怪的 Scala:

scalacOptions ++= Seq(
  "-deprecation",
  "-explaintypes",
  "-feature",
  "-language:higherKinds",
  "-unchecked",
  "-Xcheckinit",
  "-Xfatal-warnings",
  "-Xfuture",
  "-Xlint",
  "-Yno-adapted-args",
  "-Ypartial-unification",
  "-Ywarn-dead-code",
  "-Ywarn-inaccessible",
  "-Ywarn-infer-any",
  "-Ywarn-nullary-override",
  "-Ywarn-nullary-unit",
  "-Ywarn-numeric-widen",
  "-Ywarn-unused"
) ++ (
  if (scalaVersion.value.startsWith("2.11")) Seq.empty
  else Seq("-Ywarn-extra-implicit")

罗伯·诺里斯 (Rob Norris) 发表了一篇关于这些的精彩文章:
https://tpolecat.github.io/2017/04/25/scalac-flags.html

仅供引用:通用平等将被 Scala 3 中的多元平等取代，以解决您的问题:
http://dotty.epfl.ch/docs/reference/contextual/multiversal-equality.html