序列化扩展Java类的Scala类：值丢失？

Foo.java

public class Foo{
  public int i = 0;
}

斯卡拉酒吧

class Bar() extends Foo with Serializable{
  i = 1 
}

通过Josh Seureth进行序列化

REPL会话，序列化前的bar为1，序列化后为0

scala -cp .
Welcome to Scala version 2.9.1.final (Java HotSpot(TM) 64-Bit Server VM, Java 1.7.0_03).
Type in expressions to have them evaluated.
Type :help for more information.

scala> val bar = new Bar
bar: Bar = Bar@2d2ab673

scala> bar.i
res0: Int = 1

scala> :load Serialization.scala
Loading Serialization.scala...
import java.io._
defined class Serialization

scala> val serialization = new Serialization
serial: Serialization = Serialization@41a45f89

scala> serialization.write(bar)

scala> val bars = serialization.read[Bar]
bars: Bar = Bar@5a9948fd

scala> bars.i
res3: Int = 0

---

那么，为什么bars.i在这种情况下不是1呢？

这是意料之中的，我相信与Scala无关。不可序列化的超类不会被序列化（因为它们不可序列化！），因此它们的值将由默认构造函数初始化

---

如果希望以某种方式保存超类，则需要重写readObject和writeObject以手动保存状态。或者，使用更灵活的序列化解决方案，编写XML、JSON等并使用反射。

您是否尝试过使Foo可序列化？（

公共类Foo实现了可序列化的

）。我刚刚在我给出的测试用例中尝试了这一点，效果很好。然而，这是一个简化的测试用例。在我的实际问题中，我无法访问Java代码，无法添加“implements Serializable”。

scala -cp .
Welcome to Scala version 2.9.1.final (Java HotSpot(TM) 64-Bit Server VM, Java 1.7.0_03).
Type in expressions to have them evaluated.
Type :help for more information.

scala> val bar = new Bar
bar: Bar = Bar@2d2ab673

scala> bar.i
res0: Int = 1

scala> :load Serialization.scala
Loading Serialization.scala...
import java.io._
defined class Serialization

scala> val serialization = new Serialization
serial: Serialization = Serialization@41a45f89

scala> serialization.write(bar)

scala> val bars = serialization.read[Bar]
bars: Bar = Bar@5a9948fd

scala> bars.i
res3: Int = 0