Scala:"从mixin类型别名继承时,需要类类型,{{trait} {trait}找到"

tri*_*oid 2 inheritance scala mixins

我定义了一个非常常见的类型别名:

package object policy {

  type KeyGen[K] = Function0[K] with Serializable
}
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但是当我尝试继承它时:

import java.security.Key
case class FixedKeyGen(key: Key) extends KeyGen[Key] {

  override def apply(): Key = key
}
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maven编译器给了我以下错误:

[ERROR] /home/peng/git/datapassport/core/src/main/scala/com/schedule1/datapassport/policy/ValueMapping.scala:16: class type required but () => java.security.Key with Serializable found
[ERROR] case class FixedKeyGen(key: Key) extends KeyGen[Key] {
[ERROR]                                          ^
[ERROR] /home/peng/git/datapassport/core/src/main/scala/com/schedule1/datapassport/policy/ValueMapping.scala:16: com.schedule1.datapassport.policy.KeyGen[java.security.Key] does not have a constructor
[ERROR] case class FixedKeyGen(key: Key) extends KeyGen[Key] {
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这里发生了什么?

Mic*_*jac 8

我不认为你可以像这样直接扩展复合类型.也就是说,Function0[K] with Serializable它本身不是类类型.它是没有构造函数的复合类型,这是关键.在没有构造函数的情况下扩展某些东西并没有多大意义.类型别名执行与此类似的操作(请注意类型周围的括号):

case class FixedKeyGen(key: Key) extends (Function0[Key] with Serializable) {
    override def apply(): Key = key
}
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我们得到了同样的错误:

<console>:20: error: class type required but () => java.security.Key with Serializable found
       case class FixedKeyGen(key: Key) extends (Function0[Key] with Serializable) {
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这是因为Function0[Key] with Serializable不是类类型.

但是,如果我删除括号,这当然有效.没有它们,FixedKeyGen正在扩展Function0和混合Serializable.有了它们,它正试图扩展复合类型.

要解决此问题,您可能只想使用特征,而不是:

trait KeyGen[K] extends Function0[K] with Serializable

case class FixedKeyGen(key: Key) extends KeyGen[Key] {
    override def apply(): Key = key
}
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