scala – 无法将构造函数实例化为预期类型; p @ Person

发布时间：2020-12-16 18:55:21 所属栏目：安全来源：网络整理

导读：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 Rolecase ob

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",Developer)

for( person <- List(alice,bob,charlie) ) {
  person match {
    case (id,p @ Person(_,_,Manager)) => println("%s is overpaid".format(p))
    case (id,_)) => 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,Manager)) => println("%s is overpaid".format(p))
         ^
match-deep.scala:13: error: not found: value p
    case (id,Manager)) => println("%s is overpaid".format(p))
                                                                            ^
match-deep.scala:14: error: constructor cannot be instantiated to expected type;
 found   : (T1,_)) => println("%s is underpaid".format(p))
         ^
match-deep.scala:14: error: not found: value p
    case (id,_)) => println("%s is underpaid".format(p))

我在这做错了什么？

解决方法

错误信息很清楚

for( person <- List(alice,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,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,charlie).zipWithIndex ) {
     |   if(role == Manager) printf("%dth person is overpaidn",id)
     |   else printf("Something elsen")
     | }
Something else
1th person is overpaid
Something else

（编辑：李大同）

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