blocks|key|4537552|text|您正在制作HashMap本身的副本，因此更改HashMap副本不会更改原始映射(即添加或删除条目)，但是因为您存储的对象不是原语类型，所以使用给定的键检索的列表将是相同的，无论是从第一个还是第二个HashMap检索的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4537553|因此，该列表仍然只有一个副本，由两个映射引用:无论您使用哪个引用来访问它，更改列表都会更改它。|4537554|如果您希望实际的列表是一个单独的副本，则必须按照您所说的做:迭代HashMap的条目集，手动创建每个列表的副本，并将其添加到新映射中。|4537555|如果有比这更好的方法，我不知道是什么。|4537556|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|L|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|M|8|@]|9|@]|A|$]]|$1|H|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|I|$]]

You are making a copy of the HashMap itself, so changing the HashMap copy will not change the original HashMap (i.e. adding or removing entries), but because the objects you have stored are not primitive types, the List that you retrieve with a given key will be the same whether retrieved from the first or the second Map.

Thus, there is still only ONE copy of that list, referenced by both maps: changing the List changes it no matter which reference you use to access it.

If you want the actual List to be a separate copy, you will have to do as you said: iterate over the entry set of the HashMap and create a copy of each List manually, adding it to the new map as you go.

If there is a better way than that, I don't know what it is.

blocks|key|30159|text|你说得对，肤浅的拷贝不能满足你的要求。它将具有原始地图中的List的副本，但这些List将引用相同的List对象，因此一个List中对HashMap的修改将出现在另一个List中的相应HashMap中。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|30160|在Java语言中，没有为HashMap提供深度复制，因此您仍然需要遍历所有条目，并在新的HashMap中对它们执行put操作。但您也应该每次都复制一份List。如下所示：|30161|public+static+HashMap<Integer,+List<MySpecialClass>>+copy(
++++HashMap<Integer,+List<MySpecialClass>>+original)
{
++++HashMap<Integer,+List<MySpecialClass>>+copy+=+new+HashMap<Integer,+List<MySpecialClass>>();
++++for+(Map.Entry<Integer,+List<MySpecialClass>>+entry+:+original.entrySet())
++++{
++++++++copy.put(entry.getKey(),
+++++++++++//+Or+whatever+List+implementation+you'd+like+here.
+++++++++++new+ArrayList<MySpecialClass>(entry.getValue()));
++++}
++++return+copy;
}|code-block|syntax|javascript|30162|如果要修改单个MySpecialClass对象，并且不希望更改反映在复制的HashMap的List中，则还需要创建新的副本。|30163|entityMap^0|T|4|14|4|1E|4|1P|4|1V|7|2C|4|2K|7|0|C|7|18|7|1L|3|23|4|0|0|7|E|11|7|19|4|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]|$9|11|A|12|B|C]|$9|13|A|14|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|15|8|@$9|16|A|17|B|C]|$9|18|A|19|B|C]|$9|1A|A|1B|B|C]|$9|1C|A|1D|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|1E|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|1F|8|@$9|1G|A|1H|B|C]|$9|1I|A|1J|B|C]|$9|1K|A|1L|B|C]]|D|@]|E|$]]|$1|O|3|-4|5|6|7|1M|8|@]|D|@]|E|$]]]|P|$]]

You're right that a shallow copy won't meet your requirements. It will have copies of the <code>List</code>s from your original map, but those <code>List</code>s will refer to the same <code>List</code> objects, so that a modification to a <code>List</code> from one <code>HashMap</code> will appear in the corresponding <code>List</code> from the other <code>HashMap</code>.

There is no deep copying supplied for a <code>HashMap</code> in Java, so you will still have to loop through all of the entries and <code>put</code> them in the new <code>HashMap</code>. But you should also make a copy of the <code>List</code> each time also. Something like this:

<pre><code>public static HashMap&lt;Integer, List&lt;MySpecialClass&gt;&gt; copy(
 HashMap&lt;Integer, List&lt;MySpecialClass&gt;&gt; original)
{
 HashMap&lt;Integer, List&lt;MySpecialClass&gt;&gt; copy = new HashMap&lt;Integer, List&lt;MySpecialClass&gt;&gt;();
 for (Map.Entry&lt;Integer, List&lt;MySpecialClass&gt;&gt; entry : original.entrySet())
 {
 copy.put(entry.getKey(),
 // Or whatever List implementation you'd like here.
 new ArrayList&lt;MySpecialClass&gt;(entry.getValue()));
 }
 return copy;
}
</code></pre>

If you want to modify your individual <code>MySpecialClass</code> objects, and have the changes not be reflected in the <code>List</code>s of your copied <code>HashMap</code>, then you will need to make new copies of them too.

blocks|key|4537635|text|不幸的是，这确实需要迭代。但对于Java+8+streams来说，这是相当微不足道的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4537636|mapCopy+=+map.entrySet().stream()
++++.collect(Collectors.toMap(e+->+e.getKey(),+e+->+List.copyOf(e.getValue())))|code-block|syntax|javascript|4537637|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This does need iteration unfortunately. But it's pretty trivial with Java 8 streams:
<pre><code>mapCopy = map.entrySet().stream()
 .collect(Collectors.toMap(e -&gt; e.getKey(), e -&gt; List.copyOf(e.getValue())))
</code></pre>

blocks|key|4537668|text|序列化到json，然后反序列化：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4537669|Map<String,+Object>+originalMap+=+new+HashMap<>();
String+json+=+new+Gson().toJson(originalMap);
Map<String,+Object>+mapCopy+=+new+Gson().fromJson(
++++json,+new+TypeToken<Map<String,+Object>>()+{}.getType());|code-block|syntax|javascript|4537670|对于特殊的类，您可能需要write+a+custom+deserializer。|offset|length|4537671|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/google/gson/blob/master/UserGuide.md#TOC-Writing-a-Deserializer^0|0|0|C|R|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

Serialize to json and deserialize afterwards:

<pre><code>Map&lt;String, Object&gt; originalMap = new HashMap&lt;&gt;();
String json = new Gson().toJson(originalMap);
Map&lt;String, Object&gt; mapCopy = new Gson().fromJson(
 json, new TypeToken&lt;Map&lt;String, Object&gt;&gt;() {}.getType());
</code></pre>

For special classes you might need to <a href="https://github.com/google/gson/blob/master/UserGuide.md#TOC-Writing-a-Deserializer" rel="noreferrer">write a custom deserializer</a>.

blocks|key|4537708|text|只需使用Apache+Commons+Lang+-+SerializationUtils.clone()；|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4537709|如下所示：|4537710|++++Map<String,+List<MySpecialClass>>+originalMap+=+new+HashMap();
++++HashMap<String,+List<MySpecialClass>>+deepCopy+=+SerializationUtils.clone(new+HashMap<>(originalMap));|code-block|syntax|javascript|4537711|还要确保MySpecialClass实现了Serializable接口。|4537712|public+class+MySpecialClass+implements+Serializable{}|4537713|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Q|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Just use Apache Commons Lang - SerializationUtils.clone();
Something like these:
<pre><code> Map&lt;String, List&lt;MySpecialClass&gt;&gt; originalMap = new HashMap();
 HashMap&lt;String, List&lt;MySpecialClass&gt;&gt; deepCopy = SerializationUtils.clone(new HashMap&lt;&gt;(originalMap));
</code></pre>
Also make sure MySpecialClass implements Serializable interface.
<pre><code>public class MySpecialClass implements Serializable{}
</code></pre>

I need to make a copy of <code>HashMap&lt;Integer, List&lt;MySpecialClass&gt; &gt;</code> but when I change something in the copy I want the original to stay the same. i.e when I remove something from the <code>List&lt;MySpecialClass&gt;</code> from the copy it stays in the <code>List&lt;MySpecialClass&gt;</code> in the original. 

If I understand it correctly, these two methods create just shallow copy which is not what I want:

<pre><code>mapCopy = new HashMap&lt;&gt;(originalMap);
mapCopy = (HashMap) originalMap.clone();
</code></pre>

Am I right?

Is there a better way to do it than just iterate through all the keys and all the list items and copy it manually?

How to copy HashMap (not shallow copy) in Java

我需要制作HashMap<Integer, List<MySpecialClass> >的副本，但当我更改副本中的某些内容时，我希望原始版本保持不变。也就是说，当我从副本中删除List<MySpecialClass>中的内容时，它会保留在原始的List<MySpecialClass>中。如果我理解正确的话，这两种方法只...

问如何在Java中复制HashMap (非浅层复制)
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