哈希映射的序列化往返不保留顺序

Question

我注意到,最新版本的Java(1.7.0_u51)序列化和散列图的反序列化不再保留hashmap中元素的顺序.见下面的例子:

@Test
public void test() throws IOException, ClassNotFoundException {
    HashMap<String, String> map1 = new HashMap<>();
    map1.put("a1234567", "aaa");
    map1.put("b1234567", "bbb");

    System.out.println("Map1: " + map1.toString());

    byte[] serializedMap1 = objectToBytes(map1);

    System.out.println("Map1 Serialized: " + Arrays.toString(serializedMap1));

    Object map2 = bytesToObject(serializedMap1);

    System.out.println("Map2: " + map2.toString());

    byte[] serializedMap2 = objectToBytes((Serializable) map2);

    System.out.println("Map2 Serialized: " + Arrays.toString(serializedMap2));

    Object map3 = bytesToObject(serializedMap2);

    System.out.println("Map3: " + map3.toString());

    byte[] serializedMap3 = objectToBytes((Serializable) map3);

    System.out.println("Map3 Serialized: " + Arrays.toString(serializedMap3));

    Object map4 = bytesToObject(serializedMap3);

    System.out.println("Map4: " + map4.toString());

    byte[] serializedMap4 = objectToBytes((Serializable) map4);

    System.out.println("Map4 Serialized: " + Arrays.toString(serializedMap4));
}

private byte[] objectToBytes(Serializable obj) throws IOException {
    PoolByteArrayOutputStream bos = new PoolByteArrayOutputStream();
    try {
        ObjectOutputStream oos = new ObjectOutputStream(bos);
        oos.writeObject(obj);
        byte[] bytes = bos.toByteArray();
        oos.close();
        return bytes;
    } finally {
        bos.close();
    }
}

private Object bytesToObject(byte[] str) throws IOException, ClassNotFoundException {
    ByteArrayInputStream bis = new ByteArrayInputStream(str);
    ObjectInputStream ois = new ClassLoaderObjectInputStream(bis, null);

    Object obj = ois.readObject();
    ois.close();
    bis.close();
    return obj;
}

以上测试将输出:

Map1: {a1234567=aaa, b1234567=bbb}
Map1 Serialized: [-84, -19, 0, 5, 115, 114, 0, 17, 106, 97, 118, 97, 46, 117, 116, 105, 108, 46, 72, 97, 115, 104, 77, 97, 112, 5, 7, -38, -63, -61, 22, 96, -47, 3, 0, 2, 70, 0, 10, 108, 111, 97, 100, 70, 97, 99, 116, 111, 114, 73, 0, 9, 116, 104, 114, 101, 115, 104, 111, 108, 100, 120, 112, 63, 64, 0, 0, 0, 0, 0, 12, 119, 8, 0, 0, 0, 16, 0, 0, 0, 2, 116, 0, 8, 97, 49, 50, 51, 52, 53, 54, 55, 116, 0, 3, 97, 97, 97, 116, 0, 8, 98, 49, 50, 51, 52, 53, 54, 55, 116, 0, 3, 98, 98, 98, 120]
Map2: {b1234567=bbb, a1234567=aaa}
Map2 Serialized: [-84, -19, 0, 5, 115, 114, 0, 17, 106, 97, 118, 97, 46, 117, 116, 105, 108, 46, 72, 97, 115, 104, 77, 97, 112, 5, 7, -38, -63, -61, 22, 96, -47, 3, 0, 2, 70, 0, 10, 108, 111, 97, 100, 70, 97, 99, 116, 111, 114, 73, 0, 9, 116, 104, 114, 101, 115, 104, 111, 108, 100, 120, 112, 63, 64, 0, 0, 0, 0, 0, 1, 119, 8, 0, 0, 0, 2, 0, 0, 0, 2, 116, 0, 8, 98, 49, 50, 51, 52, 53, 54, 55, 116, 0, 3, 98, 98, 98, 116, 0, 8, 97, 49, 50, 51, 52, 53, 54, 55, 116, 0, 3, 97, 97, 97, 120]
Map3: {a1234567=aaa, b1234567=bbb}
Map3 Serialized: [-84, -19, 0, 5, 115, 114, 0, 17, 106, 97, 118, 97, 46, 117, 116, 105, 108, 46, 72, 97, 115, 104, 77, 97, 112, 5, 7, -38, -63, -61, 22, 96, -47, 3, 0, 2, 70, 0, 10, 108, 111, 97, 100, 70, 97, 99, 116, 111, 114, 73, 0, 9, 116, 104, 114, 101, 115, 104, 111, 108, 100, 120, 112, 63, 64, 0, 0, 0, 0, 0, 1, 119, 8, 0, 0, 0, 2, 0, 0, 0, 2, 116, 0, 8, 97, 49, 50, 51, 52, 53, 54, 55, 116, 0, 3, 97, 97, 97, 116, 0, 8, 98, 49, 50, 51, 52, 53, 54, 55, 116, 0, 3, 98, 98, 98, 120]
Map4: {b1234567=bbb, a1234567=aaa}
Map4 Serialized: [-84, -19, 0, 5, 115, 114, 0, 17, 106, 97, 118, 97, 46, 117, 116, 105, 108, 46, 72, 97, 115, 104, 77, 97, 112, 5, 7, -38, -63, -61, 22, 96, -47, 3, 0, 2, 70, 0, 10, 108, 111, 97, 100, 70, 97, 99, 116, 111, 114, 73, 0, 9, 116, 104, 114, 101, 115, 104, 111, 108, 100, 120, 112, 63, 64, 0, 0, 0, 0, 0, 1, 119, 8, 0, 0, 0, 2, 0, 0, 0, 2, 116, 0, 8, 98, 49, 50, 51, 52, 53, 54, 55, 116, 0, 3, 98, 98, 98, 116, 0, 8, 97, 49, 50, 51, 52, 53, 54, 55, 116, 0, 3, 97, 97, 97, 120]

(注意这只适用于最后7个字符相等的map键)

从上面的输出中,您可以看到每次序列化往返后订单继续更改.

我知道映射的内部顺序不能保证一致,我不依赖它,但我会假设在序列化往返之后,当映射本身没有改变时,序列化字节将是相同的.

JDK中发生了哪些具体变化导致这种情况发生？(这是JDK中的错误吗？)

有没有办法一致地为同一个hashmap获取相同的序列化字节？(不使用不同的订单保留地图)

Answer 1

HashMap没有任何可预测的顺序.因此,如果序列化改变它恰好具有的顺序,那么这不是问题.请注意,在地图中进行任何更改(添加,删除)也会更改其顺序.

如果插入顺序很重要,那么你应该使用a LinkedHashMap.