Foo.java
public class Foo{
public int i = 0;
}
Bar.scala
class Bar() extends Foo with Serializable{
i = 1
}
通过Josh Seureth https://stackoverflow.com/a/3442574/390708进行序列化
import java.io._
class Serialization{
def write(x : AnyRef) {
val output = new ObjectOutputStream(new FileOutputStream("test.obj"))
output.writeObject(x)
output.close()
}
def read[A] = {
val input = new ObjectInputStream(new FileInputStream("test.obj"))
val obj = input.readobject()
input.close()
obj.asInstanceOf[A]
}
}
REPL会话,序列化之前的条形码是1,但是之后是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,为什么在这种情况下不是bars.i 1?
解决方法:
这是预期的,我相信与Scala无关.非序列化超类未被序列化(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.
如果您想以某种方式保存超类,则需要覆盖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来编写XML,JSON等并使用反射.
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