Scala 2.12反序列化Scala 2.11序列化的Symbol兼容性问题

问题场景 #

代码示例：

import java.io.{FileInputStream, FileOutputStream, ObjectInputStream, ObjectOutputStream}
object SymbolSerializeDemo {
  def main(args: Array[String]): Unit = {
    val fileName = "file.ser"
    val symbolCheck: Symbol = Symbol("someSymbol")
    //serializeToFile(symbolCheck, fileName)
    deserializeFromFile(fileName)
  }
  private def serializeToFile(input: Symbol, fileName: String): Unit = {
    try {
      val file: FileOutputStream = new FileOutputStream(fileName)
      val out: ObjectOutputStream = new ObjectOutputStream(file)
      out.writeObject(input)
    }
  }
  private def deserializeFromFile(fileName: String): Unit = {
    try {
      val file: FileInputStream = new FileInputStream(fileName)
      val input: ObjectInputStream = new ObjectInputStream(file)
      val output = input.readObject.asInstanceOf[Symbol]
      println("Symbol after deseralization " + output.name)
    }
  }
}

在Scala 2.12中反序列化Scala 2.11序列化的scala.Symbol对象时，抛出异常：

java.io.InvalidClassException: scala.Symbol; local class incompatible: stream classdesc serialVersionUID = 2966401305346518859, local class serialVersionUID = 6865603221856321286

尝试为自定义类及Symbol添加serialVersionUID无效，因为Symbol是Scala标准库类，无法修改其内部定义。

解决方案 #

方案1：替换序列化逻辑（推荐）

Symbol的核心是其name字符串，序列化时直接存储字符串，反序列化时重新创建Symbol对象，从根源避免版本兼容问题：

修改后的代码：

import java.io.{FileInputStream, FileOutputStream, ObjectInputStream, ObjectOutputStream}

object SymbolSerializeDemo {
  def main(args: Array[String]): Unit = {
    val fileName = "file.ser"
    val symbolCheck: Symbol = Symbol("someSymbol")
    // 序列化时存字符串
    serializeSymbolByName(symbolCheck, fileName)
    // 反序列化时重建Symbol
    val restoredSymbol = deserializeSymbolFromName(fileName)
    println("Symbol after deserialization: " + restoredSymbol.name)
  }

  private def serializeSymbolByName(symbol: Symbol, fileName: String): Unit = {
    val file = new FileOutputStream(fileName)
    val out = new ObjectOutputStream(file)
    try {
      out.writeUTF(symbol.name)
    } finally {
      out.close()
      file.close()
    }
  }

  private def deserializeSymbolFromName(fileName: String): Symbol = {
    val file = new FileInputStream(fileName)
    val in = new ObjectInputStream(file)
    try {
      val name = in.readUTF()
      Symbol(name)
    } finally {
      in.close()
      file.close()
    }
  }
}

方案2：修复已有序列化文件的反序列化

如果已经有大量用Scala 2.11序列化的Symbol文件，需要在Scala 2.12中读取，可以自定义ObjectInputStream，重写resolveClass方法，忽略serialVersionUID的差异：

import java.io.{FileInputStream, ObjectInputStream, ObjectStreamClass}

private class CompatibleSymbolInputStream(filePath: String) extends ObjectInputStream(new FileInputStream(filePath)) {
  override def resolveClass(desc: ObjectStreamClass): Class[_] = {
    if (desc.getName == classOf[Symbol].getName) {
      // 强制使用本地的Symbol类，忽略serialVersionUID差异
      classOf[Symbol]
    } else {
      super.resolveClass(desc)
    }
  }
}

// 修改原反序列化方法
private def deserializeFromFile(fileName: String): Unit = {
  val in = new CompatibleSymbolInputStream(fileName)
  try {
    val output = in.readObject().asInstanceOf[Symbol]
    println("Symbol after deserialization: " + output.name)
  } finally {
    in.close()
  }
}

注意：这种方法依赖Symbol类的内部结构在2.11到2.12之间没有发生破坏性变化（实际上Scala 2.11到2.12的Symbol内部主要是serialVersionUID变化，字段还是只有name），所以可以正常读取。但如果后续版本Symbol结构变化，这种方法可能失效。

内容的提问来源于stack exchange，提问作者user3771712