scala – 无法将构造函数实例化为预期类型; p @ Person

我使用scala版本:Scala代码运行版本2.9.2-未知 – 未知 – 版权所有2002-2011,LAMP / EPFL

我在这里尝试深度大小写匹配构造:http://ofps.oreilly.com/titles/9780596155957/RoundingOutTheEssentials.html,代码如下match-deep.scala:

class Role
case object Manager extends Role
case object Developer extends Role

case class Person(name:String, age: Int, role: Role)

val alice = new Person("Alice", 25, Developer)
val bob = new Person("Bob", 32, Manager)
val charlie = new Person("Charlie", 32, Developer)

for( person <- List(alice, bob, charlie) ) {
  person match {
    case (id, p @ Person(_, _, Manager)) => println("%s is overpaid".format(p))
    case (id, p @ Person(_, _, _)) => println("%s is underpaid".format(p))
  }
}

我收到以下错误:

match-deep.scala:13: error: constructor cannot be instantiated to expected type;
 found   : (T1, T2)
 required: this.Person
    case (id, p @ Person(_, _, Manager)) => println("%s is overpaid".format(p))
         ^
match-deep.scala:13: error: not found: value p
    case (id, p @ Person(_, _, Manager)) => println("%s is overpaid".format(p))
                                                                            ^
match-deep.scala:14: error: constructor cannot be instantiated to expected type;
 found   : (T1, T2)
 required: this.Person
    case (id, p @ Person(_, _, _)) => println("%s is underpaid".format(p))
         ^
match-deep.scala:14: error: not found: value p
    case (id, p @ Person(_, _, _)) => println("%s is underpaid".format(p))

我在这做错了什么?

最佳答案
错误信息很清楚

for( person <- List(alice, bob, charlie) ) {
  person match {
    case p @ Person(_, _, Manager) => println("%s is overpaid".format(p.toString))
    case p @ Person(_, _, _) => println("%s is underpaid".format(p.toString))
  }
}

这是一个做同样事情的简短方法:

for(p @ Person(_, _, role) <- List(alice, bob, charlie) ) {
  if(role == Manager) println("%s is overpaid".format(p.toString))
  else println("%s is underpaid".format(p.toString))
}

编辑
我不确定你想要的id的真正含义是什么,我想它是列表中人物的索引.开始了:

scala> for((p@Person(_,_,role), id) <- List(alice, bob, charlie).zipWithIndex ) {
     |   if(role == Manager) printf("%dth person is overpaid\n", id)
     |   else printf("Something else\n")
     | }
Something else
1th person is overpaid
Something else

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