blocks|key|709252|text|当我自己面对同样的问题时也遇到了这个问题。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|709253|声明arraylist1+=+arraylist2会将这两个列表设置为指向同一位置，因此，如果您更改其中一个数据，那么两个列表将始终保持不变。|709254|为了将值复制到一个独立的列表中，我只使用foreach来复制内容：|709255|ArrayList+list1+=+new+ArrayList();
ArrayList+list2+=+new+ArrayList();|code-block|syntax|javascript|709256|以您当前的方式填充list1。|709257|foreach(<type>+obj+in+list1)
{
++++list2.Add(obj);
}|709258|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|S|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|T|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|H|7|V|8|@]|9|@]|A|$I|J]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

Came across this while facing the same issue myself.

Saying arraylist1 = arraylist2 sets them both to point at the same place so if you alter either the data alters and thus both lists always stay the same.

To copy values into an independent list I just used foreach to copy the contents:

<pre><code>ArrayList list1 = new ArrayList();
ArrayList list2 = new ArrayList();
</code></pre>

fill list1 in whatever way you currently are.

<pre><code>foreach(&lt;type&gt; obj in list1)
{
 list2.Add(obj);
}
</code></pre>

blocks|key|709316|text|支持将oldList复制到名为newList的新ArrayList对象中|type|unstyled|depth|inlineStyleRanges|entityRanges|data|709317|ArrayList<Object>+newList+=+new+ArrayList<>()+;

for+(int+i+=+0+;+i<oldList.size();i%2B%2B){
++++newList.add(oldList.get(i))+;
}|code-block|syntax|javascript|709318|这两个列表是独立的，其中一个列表的更改不会反映到另一个列表中。|709319|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|$]]

Supopose you want to copy oldList into a new ArrayList object called newList

<pre><code>ArrayList&lt;Object&gt; newList = new ArrayList&lt;&gt;() ;

for (int i = 0 ; i&lt;oldList.size();i++){
 newList.add(oldList.get(i)) ;
}
</code></pre>

These two lists are indepedant, changes to one are not reflected to the other one.

blocks|key|493400|text|您需要对单个对象执行clone()。Constructor和其他方法执行浅层复制。您可以尝试Collections.copy方法。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|493401|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/6/docs/api/java/util/Collections.html#copy%2528java.util.List,%2520java.util.List%2529^0|A|7|I|B|1A|G|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@$9|O|A|P|B|C]|$9|Q|A|R|B|C]]|D|@$9|S|A|T|1|U]]|E|$]]|$1|F|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|G|$H|$5|I|J|K|E|$L|M]]]]

You need to <code>clone()</code> the individual object. <code>Constructor</code> and other methods perform shallow copy. You may try <a href="http://docs.oracle.com/javase/6/docs/api/java/util/Collections.html#copy%28java.util.List,%20java.util.List%29" rel="nofollow">Collections.copy</a> method.

blocks|key|493605|text|直接复制原始列表的方法是将一个列表中的所有元素添加到另一个列表中。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|493606|ArrayList<Object>+originalList+=+new+ArrayList<Object>();
ArrayList<Object>+duplicateList+=+new+ArrayList<Object>();

for(Object+o+:+originalList)+{
++++duplicateList.add(o);
}|code-block|syntax|javascript|493607|现在，如果您对originalList进行任何更改，都不会影响duplicateList。|493608|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|$]]

Straightforward way to make deep copy of original list is to add all element from one list to another list.

<pre><code>ArrayList&lt;Object&gt; originalList = new ArrayList&lt;Object&gt;();
ArrayList&lt;Object&gt; duplicateList = new ArrayList&lt;Object&gt;();

for(Object o : originalList) {
 duplicateList.add(o);
}
</code></pre>

Now If you make any changes to originalList it will not impact duplicateList.

blocks|key|493431|text|要将一个列表复制到另一个列表中，可以使用名为Collection.copy(myObject+myTempObject).now的方法。执行这行代码后，您可以看到myObject中的所有列表值。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|493432|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|$]]

to copy one list into the other list, u can use the method called
Collection.copy(myObject myTempObject).now after executing these line of code u can see all the list values in the myObject.

blocks|key|493695|text|以下是将所有对象从一个arrayList复制到另一个all的解决方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|493696|+ArrayList<Object>+myObject+=+new+ArrayList<Object>();
ArrayList<Object>+myTempObject+=+new+ArrayList<Object>();

myObject.addAll(myTempObject.subList(0,+myTempObject.size()));|code-block|syntax|javascript|493697|subList旨在返回一个包含一系列数据的列表。因此您可以复制整个arrayList或其中的一部分。|493698|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|$]]

Here is a workaround to copy all the objects from one arrayList to another:

<pre><code> ArrayList&lt;Object&gt; myObject = new ArrayList&lt;Object&gt;();
ArrayList&lt;Object&gt; myTempObject = new ArrayList&lt;Object&gt;();

myObject.addAll(myTempObject.subList(0, myTempObject.size()));
</code></pre>

subList is intended to return a List with a range of data. so you can copy the whole arrayList or part of it.

blocks|key|709509|text|假设您有两个String类型数组列表。喜欢|type|unstyled|depth|inlineStyleRanges|entityRanges|data|709510|ArrayList<String>+firstArrayList+;//This+array+list+is+not+having+any+data.

ArrayList<String>+secondArrayList+=+new+ArrayList<>();//Having+some+data.|code-block|syntax|javascript|709511|现在我们必须像这样将第二个数组的数据复制到第一个数组列表中，|709512|firstArrayList+=+new+ArrayList<>(secondArrayList+);|709513|完成了！！|709514|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Suppose you have two arraylist of String type .
Like 

<pre><code>ArrayList&lt;String&gt; firstArrayList ;//This array list is not having any data.

ArrayList&lt;String&gt; secondArrayList = new ArrayList&lt;&gt;();//Having some data.
</code></pre>

Now we have to copy the data of second array to first arraylist like this,

<pre><code>firstArrayList = new ArrayList&lt;&gt;(secondArrayList );
</code></pre>

Done!!

blocks|key|709535|text|最简单的方法是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|709536|ArrayList<Object>+myObject+=+new+ArrayList<Object>();
//+fill+up+data+here
ArrayList<Object>+myTempObject+=+new+ArrayList(myObject);|code-block|syntax|javascript|709537|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|$]]

The simplest way is: 
<pre><code>ArrayList&lt;Object&gt; myObject = new ArrayList&lt;Object&gt;();
// fill up data here
ArrayList&lt;Object&gt; myTempObject = new ArrayList(myObject);
</code></pre>

I have some data structures, and I would like to use one as a temporary, and another as not temporary.

<pre><code>ArrayList&lt;Object&gt; myObject = new ArrayList&lt;Object&gt;();
ArrayList&lt;Object&gt; myTempObject = new ArrayList&lt;Object&gt;();


//fill myTempObject here
....

//make myObject contain the same values as myTempObject
myObject = myTempObject;

//free up memory by clearing myTempObject
myTempObject.clear();
</code></pre>

now the problem with this of course is that <code>myObject</code> is really just pointing to <code>myTempObject</code>, and so once <code>myTempObject</code> is cleared, so is <code>myObject</code>.

How do I retain the values from <code>myTempObject</code> in <code>myObject</code> using java?

How do I copy the contents of one ArrayList into another?

数据结构

我有一些数据结构，我想使用一个作为临时的，另一个作为非临时的。ArrayList<Object> myObject = new ArrayList<Object>();ArrayList<Object> myTempObject = new ArrayList<Object>();//fill myTempObject here....//make myObject contain the sam

问如何将一个ArrayList的内容复制到另一个？
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