可以解构导致未来的错误吗？

Question

最初我们有一个类似的代码

class Person(firstName: String, lastName: String)

val (firstName, lastName) = person

有人交换领域

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名称.只要typer类型检查,一切都很好.你可以做的是额外的类型:

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

class Person(firstName: FirstName, lastName: LastName)

现在交换类型将导致编译时错误.

如果您不希望分配其他对象的额外开销,请查看无形标记类型:

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指出使用价值类:

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