这只是一个被接受的答案的优化代码形式，以防有人想在生产中使用它:

    public static void byteArrayOps() throws IOException, ClassNotFoundException{

    String str="123";
     byte[] yourBytes = null;

    // Convert to byte[]

    try(ByteArrayOutputStream bos = new ByteArrayOutputStream();
            ObjectOutputStream out =  new ObjectOutputStream(bos);) {


      out.writeObject(str);
      out.flush();
      yourBytes = bos.toByteArray();

    } finally {

    }

    // convert back to Object

    try(ByteArrayInputStream bis = new ByteArrayInputStream(yourBytes);
            ObjectInput in = new ObjectInputStream(bis);) {

      Object o = in.readObject(); 

    } finally {

    }




}

如果你使用Java >= 7，你可以使用try和资源改进接受的解决方案:

private byte[] convertToBytes(Object object) throws IOException {
    try (ByteArrayOutputStream bos = new ByteArrayOutputStream();
         ObjectOutputStream out = new ObjectOutputStream(bos)) {
        out.writeObject(object);
        return bos.toByteArray();
    } 
}

反过来说:

private Object convertFromBytes(byte[] bytes) throws IOException, ClassNotFoundException {
    try (ByteArrayInputStream bis = new ByteArrayInputStream(bytes);
         ObjectInputStream in = new ObjectInputStream(bis)) {
        return in.readObject();
    } 
}

这只是一个被接受的答案的优化代码形式，以防有人想在生产中使用它:

    public static void byteArrayOps() throws IOException, ClassNotFoundException{

    String str="123";
     byte[] yourBytes = null;

    // Convert to byte[]

    try(ByteArrayOutputStream bos = new ByteArrayOutputStream();
            ObjectOutputStream out =  new ObjectOutputStream(bos);) {


      out.writeObject(str);
      out.flush();
      yourBytes = bos.toByteArray();

    } finally {

    }

    // convert back to Object

    try(ByteArrayInputStream bis = new ByteArrayInputStream(yourBytes);
            ObjectInput in = new ObjectInputStream(bis);) {

      Object o = in.readObject(); 

    } finally {

    }




}

最好的方法是使用Apache Commons Lang中的SerializationUtils。

序列化:

byte[] data = SerializationUtils.serialize(yourObject);

反序列化:

YourObject yourObject = SerializationUtils.deserialize(data)

如前所述，这需要Commons Lang库。它可以使用Gradle导入:

compile 'org.apache.commons:commons-lang3:3.5'

Maven:

<!-- https://mvnrepository.com/artifact/org.apache.commons/commons-lang3 -->
<dependency>
    <groupId>org.apache.commons</groupId>
    <artifactId>commons-lang3</artifactId>
    <version>3.5</version>
</dependency>

Jar文件

这里提到了更多的方法

或者，可以导入整个集合。参考这个连结

准备要发送的字节数组:

ByteArrayOutputStream bos = new ByteArrayOutputStream();
ObjectOutputStream out = null;
try {
  out = new ObjectOutputStream(bos);   
  out.writeObject(yourObject);
  out.flush();
  byte[] yourBytes = bos.toByteArray();
  ...
} finally {
  try {
    bos.close();
  } catch (IOException ex) {
    // ignore close exception
  }
}

从字节数组创建一个对象:

ByteArrayInputStream bis = new ByteArrayInputStream(yourBytes);
ObjectInput in = null;
try {
  in = new ObjectInputStream(bis);
  Object o = in.readObject(); 
  ...
} finally {
  try {
    if (in != null) {
      in.close();
    }
  } catch (IOException ex) {
    // ignore close exception
  }
}

另一个有趣的方法来自com.fasterxml.jackson.databind.ObjectMapper

byte[] data = new ObjectMapper().writeValueAsBytes(JAVA_OBJECT_HERE)

Maven的依赖

<dependency>
    <groupId>com.fasterxml.jackson.core</groupId>
    <artifactId>jackson-databind</artifactId>
</dependency>