针对’应该包含’的最新自定义匹配器

import org.scalactic.Equality
import org.scalatest.{FreeSpec,Matchers}

final class Test extends FreeSpec with Matchers {

  case class Person(name: String)

  val people = List(Person("Alice"),Person("Eve"))

  val namesBeingEqual = MappingEquality[Person,String](p => p.name)

  "test should pass" in {
    (people should contain theSameElementsAs List("Alice","Eve"))(
      decided by namesBeingEqual)
  }

  "test should fail" in {
    (people should contain theSameElementsAs List("Alice","Bob"))(
      decided by namesBeingEqual)
  }

  case class MappingEquality[S,T](map: S => T) extends Equality[S] {

    override def areEqual(s: S,b: Any): Boolean = b match {
      case t: T => map(s) == t
      case _    => false
    }
  }
}

我不确定我是否会说这是一个好主意,因为它并不像人们期望的任何称为Equality的行为那样完美,但它有效.

您甚至可以通过隐式转换将其添加到after之后获取您建议的beingMapped语法：

implicit class AfterExtensions(aft: TheAfterWord) {
    def beingMapped[S,T](map: S => T): Equality[S] = MappingEquality(map)
    }
  }

我确实试过通过Uniformity trait来实现它,它有类似的方法涉及Any,但遇到了问题,因为规范化是错误的方法：我可以从你的例子创建Uniformity [String]对象,但不是Uniformity [人]一个. (原因是有一个规范化的方法返回用于构造Equality对象的泛型类型,这意味着为了将字符串与字符串进行比较,左侧输入必须是一个字符串.)这意味着唯一的写入方式与正常情况下的预期vs实际值相反：

"test should succeed" in {
    val mappedToName = MappingUniformity[Person,String](person => person.name)
    (List("Alice","Eve") should contain theSameElementsAs people)(
      after being mappedToName)
  }

  case class MappingUniformity[S,T](map: S => T) extends Uniformity[T] {

    override def normalizedOrSame(b: Any): Any = b match {
      case s: S => map(s)
      case t: T => t
    }

    override def normalizedCanHandle(b: Any): Boolean =
      b.isInstanceOf[S] || b.isInstanceOf[T]

    override def normalized(s: T): T = s
  }

绝对不是你通常想写这个的方式.