解构将来会导致错误吗？

Question

最初我们有这样的代码

class Person(firstName: String, lastName: String)

val (firstName, lastName) = person

有人交换字段

class Person(lastName: String, firstName: String)

val (firstName, lastName) = person

编译器很高兴！

Answer 1

编译器无法知道您交换了字段的顺序，因为分析不会查找参数 name。只要打字机类型检查，一切都很好。您可以做的是额外级别的类型：

case class FirstName(name: String)
case class LastName(name: String)

class Person(firstName: FirstName, lastName: LastName)

现在交换类型会导致编译时错误。

如果您不想分配额外对象的额外开销，请查看 shapeless tagged types:

import shapeless.tag.@@

trait FirstNameTag
trait LastNameTag

type FirstName = String @@ FirstNameTag
type LastName = String @@ LastNameTag

case class Person(fn: FirstName, ln: LastName)

现在：

import shapeless.tag

val firstName: FirstName = tag[FirstNameTag][String]("yuv")
val lastName: LastName = tag[LastName][String]("itz")

val person = Person(firstName, lastName)

或者@Ren 指出使用 value classes:

case class FirstName(name: String) extends AnyVal
case class LastName(name: String) extends AnyVal

解构将来会导致错误吗？

Can destructuring cause bugs in future?

scala

destructuring