blocks|key|1554589|text|ArrayList+newArrayList+=+(ArrayList)+oldArrayList.clone();|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1554590|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>ArrayList newArrayList = (ArrayList) oldArrayList.clone();
</code></pre>

blocks|key|1769732|text|你为什么要克隆呢？创建一个新的列表通常更有意义。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1769733|List<String>+strs;
...
List<String>+newStrs+=+new+ArrayList<>(strs);|code-block|syntax|javascript|1769734|任务完成了。|1769735|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Why would you want to clone? Creating a new list usually makes more sense.

<pre><code>List&lt;String&gt; strs;
...
List&lt;String&gt; newStrs = new ArrayList&lt;&gt;(strs);
</code></pre>

Job done.

blocks|key|1554763|text|这是我用来做这个的代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1554764|ArrayList+copy+=+new+ArrayList+(original.size());
Collections.copy(copy,+original);|code-block|syntax|javascript|1554765|希望对你很有用|1554766|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

This is the code I use for that:

<pre><code>ArrayList copy = new ArrayList (original.size());
Collections.copy(copy, original);
</code></pre>

Hope is usefull for you

blocks|key|1554686|text|需要注意的是，Object.clone()有一些主要问题，在大多数情况下都不鼓励使用它。请参阅Joshua+Bloch的"Effective+Java“中的第11项，以获得完整的答案。我相信您可以在原始类型数组上安全地使用Object.clone()，但除此之外，您还需要明智地正确使用和覆盖克隆。您最好定义一个复制构造函数或一个根据语义显式克隆对象的静态工厂方法。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1554687|entityMap|0|LINK|mutability|MUTABLE|url|http://java.sun.com/docs/books/effective/^0|1P|E|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

Be advised that Object.clone() has some major problems, and its use is discouraged in most cases. Please see Item 11, from "<a href="http://java.sun.com/docs/books/effective/" rel="noreferrer">Effective Java</a>" by Joshua Bloch for a complete answer. I believe you can safely use Object.clone() on primitive type arrays, but apart from that you need to be judicious about properly using and overriding clone. You are probably better off defining a copy constructor or a static factory method that explicitly clones the object according to your semantics.

blocks|key|1769913|text|在Java+8中，可以使用流来克隆它。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1769914|import+static+java.util.stream.Collectors.toList;|code-block|syntax|javascript|1769915|..。|1769916|List<AnObject>+clone+=+myList.stream().collect(toList());|1769917|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

With Java 8 it can be cloned with a stream.

<pre><code>import static java.util.stream.Collectors.toList;
</code></pre>

...

<pre><code>List&lt;AnObject&gt; clone = myList.stream().collect(toList());
</code></pre>

blocks|key|1554611|text|我发现使用addAll工作得很好。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1554612|ArrayList<String>+copy+=+new+ArrayList<String>();
copy.addAll(original);|code-block|syntax|javascript|1554613|使用括号而不是泛型语法|1554614|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

I find using addAll works fine.

<pre><code>ArrayList&lt;String&gt; copy = new ArrayList&lt;String&gt;();
copy.addAll(original);
</code></pre>

parentheses are used rather than the generics syntax

blocks|key|1769943|text|如果需要这样做，以便能够在getter中返回列表，最好这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1769944|ImmutableList.copyOf(list);|code-block|syntax|javascript|1769945|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|$]]

If you want this in order to be able to return the List in a getter it would be better to do:

<pre><code>ImmutableList.copyOf(list);
</code></pre>

blocks|key|1554907|text|List<String>+shallowClonedList+=+new+ArrayList<>(listOfStrings);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1554908|请记住，这只是一个浅的，而不是一个深的副本，即。您将获得一个新列表，但条目是相同的。这对于简单的字符串来说是没有问题的。当列表条目本身是对象时，Get会更加棘手。|unstyled|1554909|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>List&lt;String&gt; shallowClonedList = new ArrayList&lt;&gt;(listOfStrings);
</code></pre>

Keep in mind that this is only a shallow not a deep copy, ie. you get a new list, but the entries are the same. This is no problem for simply strings. Get's more tricky when the list entries are objects themself.

blocks|key|1554807|text|要克隆像java.util.List这样的泛型接口，您只需对其进行强制转换。下面是一个示例：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1554808|List+list+=+new+ArrayList();
List+list2+=+((List)+(+(ArrayList)+list).clone());|code-block|syntax|javascript|1554809|这有点棘手，但它是有效的，如果您被限制为返回一个List接口，那么您之后的任何人都可以随时实现您的list。|1554810|我知道这个答案接近最终答案，但我的答案是在您使用List+-the泛型父级而不是ArrayList时如何做到这一点|1554811|entityMap^0|4|E|0|0|O|4|0|O|4|14|9|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|X|8|@$9|Y|A|Z|B|C]|$9|10|A|11|B|C]]|D|@]|E|$]]|$1|O|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|P|$]]

To clone a generic interface like <code>java.util.List</code> you will just need to cast it. here you are an example:

<pre><code>List list = new ArrayList();
List list2 = ((List) ( (ArrayList) list).clone());
</code></pre>

It is a bit tricky, but it works, if you are limited to return a <code>List</code> interface, so anyone after you can implement your list whenever he wants.

I know this answer is close to the final answer, but my answer answers how to do all of that while you are working with <code>List</code> -the generic parent- not <code>ArrayList</code>

blocks|key|1554635|text|这也应该是可行的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1554636|ArrayList<String>+orig+=+new+ArrayList<String>();
ArrayList<String>+copy+=+(ArrayList<String>)+orig.clone()|code-block|syntax|javascript|1554637|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 should also work:

<pre><code>ArrayList&lt;String&gt; orig = new ArrayList&lt;String&gt;();
ArrayList&lt;String&gt; copy = (ArrayList&lt;String&gt;) orig.clone()
</code></pre>

blocks|key|1554709|text|在克隆ArrayLists时要非常小心。在java中克隆是很肤浅的。这意味着它只会克隆数组列表本身，而不是它的成员。因此，如果您有一个ArrayList+X1并将其克隆到X2中，则X2中的任何更改也将在X1中体现出来，反之亦然。在克隆时，您只会生成一个新的ArrayList，其中包含指向原始a中相同元素的指针。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1554710|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Be very careful when cloning ArrayLists. Cloning in java is shallow. This means that it will only clone the Arraylist itself and not its members. So if you have an ArrayList X1 and clone it into X2 any change in X2 will also manifest in X1 and vice-versa. When you clone you will only generate a new ArrayList with pointers to the same elements in the original.

blocks|key|1769646|text|ArrayList+first+=+new+ArrayList+();
ArrayList+copy+=+(ArrayList)+first.clone+();|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1769647|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>ArrayList first = new ArrayList ();
ArrayList copy = (ArrayList) first.clone ();
</code></pre>

blocks|key|1769815|text|我不是java专业人士，但我也有同样的问题，我试着用这种方法来解决。(假设T有一个复制构造函数)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1769816|+public+static+<T+extends+Object>+List<T>+clone(List<T>+list)+{
++++++try+{
+++++++++++List<T>+c+=+list.getClass().newInstance();
+++++++++++for(T+t:+list)+{
+++++++++++++T+copy+=+(T)+t.getClass().getDeclaredConstructor(t.getclass()).newInstance(t);
+++++++++++++c.add(copy);
+++++++++++}
+++++++++++return+c;
++++++}+catch(Exception+e)+{
+++++++++++throw+new+RuntimeException("List+cloning+unsupported",e);
++++++}
}|code-block|syntax|javascript|1769817|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|$]]

I am not a java professional, but I have the same problem and I tried to solve by this method. (It suppose that T has a copy constructor).

<pre><code> public static &lt;T extends Object&gt; List&lt;T&gt; clone(List&lt;T&gt; list) {
 try {
 List&lt;T&gt; c = list.getClass().newInstance();
 for(T t: list) {
 T copy = (T) t.getClass().getDeclaredConstructor(t.getclass()).newInstance(t);
 c.add(copy);
 }
 return c;
 } catch(Exception e) {
 throw new RuntimeException("List cloning unsupported",e);
 }
}
</code></pre>

I have an <code>ArrayList&lt;String&gt;</code> that I'd like to return a copy of. <code>ArrayList</code> has a clone method which has the following signature:

<pre><code>public Object clone()
</code></pre>

After I call this method, how do I cast the returned Object back to <code>ArrayList&lt;String&gt;</code>?

How do I clone a generic List in Java?

我有一个ArrayList<String>，我想退回一份。ArrayList有一个clone方法，其签名如下：public Object clone()调用此方法后，如何将返回的对象转换回ArrayList<String>

问如何在Java中克隆泛型列表？
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