如何在scala中实现接口Serializable?

本文介绍了如何在scala中实现接口Serializable?的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我有像这样的 Scala 类:

I have scala class like:

@Entity("users") class User(@Required val cid: String, val isAdmin: Boolean = false, @Required val dateJoined: Date = new Date() ) { @Id var id: ObjectId = _ @Reference val foos = new ArrayList[Foo] }

如果它是一个 Java 类，我会简单地放置实现 java.io.Serializable 但这在 Scala 中不起作用.上面声明的 foos 是私有的还是公共的?

If it was a Java class I would simply put implements java.io.Serializable but this does not work in scala. Also is foos as declared above is private or public?

推荐答案

scala 2.9.x 也有一个名为 Serializable 的接口，你可以扩展或混合它.在 2.9.x 之前，@serializable 是唯一的选择.

scala 2.9.x also have an interface named Serializable, you may extends or mixin this. before 2.9.x the @serializable is the only choice.