blocks|key|1050375|text|这样做是不安全的！|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1050376|想象一下，如果你有：|1050377|ArrayList<Object>+list+=+new+ArrayList<Object>();
list.add(new+Employee("Jonh"));
list.add(new+Car("BMW","M3"));
list.add(new+Chocolate("Twix"));|code-block|syntax|javascript|1050378|将这些对象的列表转换为任何类型都没有意义。|1050379|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|O|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

It's not safe to do that! 
Imagine if you had: 

<pre><code>ArrayList&lt;Object&gt; list = new ArrayList&lt;Object&gt;();
list.add(new Employee("Jonh"));
list.add(new Car("BMW","M3"));
list.add(new Chocolate("Twix"));
</code></pre>

It wouldn't make sense to convert the list of those Objects to any type.

blocks|key|1050593|text|使用Java+8，您可以做到：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1050594|List<Object>+list+=+...;
List<String>+strList+=+list.stream()
+++++++++++++++++++++++++++.map(+Object::toString+)
+++++++++++++++++++++++++++.collect(+Collectors.toList()+);|code-block|syntax|javascript|1050595|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|$]]

Using Java 8 you can do:

<pre><code>List&lt;Object&gt; list = ...;
List&lt;String&gt; strList = list.stream()
 .map( Object::toString )
 .collect( Collectors.toList() );
</code></pre>

blocks|key|836004|text|您可以使用通配符执行此操作，如下所示|type|unstyled|depth|inlineStyleRanges|entityRanges|data|836005|ArrayList<String>+strList+=+(ArrayList<String>)(ArrayList<?>)(list);|code-block|syntax|javascript|836006|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|$]]

You can use wildcard to do this as following 

<pre><code>ArrayList&lt;String&gt; strList = (ArrayList&lt;String&gt;)(ArrayList&lt;?&gt;)(list);
</code></pre>

blocks|key|1050539|text|如果你想用下流的方式来做，试试这个。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1050540|@SuppressWarnings("unchecked")
public+ArrayList<String>+convert(ArrayList<Object>+a)+{
+++return+(ArrayList)+a;
}|code-block|syntax|javascript|1050541|优点:在这里，您无需迭代所有对象，从而节省了时间。|1050542|缺点:可能会在你的脚上留下一个洞。|1050543|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]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

If you want to do it the dirty way, try this.

<pre><code>@SuppressWarnings("unchecked")
public ArrayList&lt;String&gt; convert(ArrayList&lt;Object&gt; a) {
 return (ArrayList) a;
}
</code></pre>

Advantage: here you save time by not iterating over all objects.

Disadvantage: may produce a hole in your foot.

blocks|key|1050436|text|使用guava|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1050437|List<String>+stringList=Lists.transform(list,new+Function<Object,String>(){
++++@Override
++++public+String+apply(Object+arg0)+{
++++++++if(arg0!=null)
++++++++++++return+arg0.toString();
++++++++else
++++++++++++return+"null";
++++}
});|code-block|syntax|javascript|1050438|entityMap|0|LINK|mutability|MUTABLE|url|http://code.google.com/p/guava-libraries/^0|2|5|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

Using <a href="http://code.google.com/p/guava-libraries/" rel="noreferrer">guava</a>:

<pre><code>List&lt;String&gt; stringList=Lists.transform(list,new Function&lt;Object,String&gt;(){
 @Override
 public String apply(Object arg0) {
 if(arg0!=null)
 return arg0.toString();
 else
 return "null";
 }
});
</code></pre>

blocks|key|835732|text|下面是使用Guava的另一种选择|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|835733|List<Object>+lst+...++++
List<String>+ls+=+Lists.transform(lst,+Functions.toStringFunction());|code-block|syntax|javascript|835734|entityMap|0|LINK|mutability|MUTABLE|url|https://code.google.com/p/guava-libraries/^0|5|5|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

Here is another alternative using <a href="https://code.google.com/p/guava-libraries/" rel="noreferrer">Guava</a> 

<pre><code>List&lt;Object&gt; lst ... 
List&lt;String&gt; ls = Lists.transform(lst, Functions.toStringFunction());
</code></pre>

blocks|key|835657|text|您的代码ArrayList<String>+list2+=+(String)list;无法编译，因为list2不是String类型。但这并不是唯一的问题。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|835658|entityMap^0|4|13|1E|5|1L|6|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]|$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|O|8|@]|D|@]|E|$]]]|G|$]]

Your code <code>ArrayList&lt;String&gt; list2 = (String)list;</code> does not compile because <code>list2</code> is not of type <code>String</code>. But that is not the only problem.

blocks|key|835850|text|使用Java+8+lambda：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|835851|ArrayList<Object>+obj+=+new+ArrayList<>();
obj.add(1);
obj.add("Java");
obj.add(3.14);

ArrayList<String>+list+=+new+ArrayList<>();
obj.forEach((xx)+->+list.add(String.valueOf(xx)));|code-block|syntax|javascript|835852|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|$]]

Using Java 8 lambda:

<pre><code>ArrayList&lt;Object&gt; obj = new ArrayList&lt;&gt;();
obj.add(1);
obj.add("Java");
obj.add(3.14);

ArrayList&lt;String&gt; list = new ArrayList&lt;&gt;();
obj.forEach((xx) -&gt; list.add(String.valueOf(xx)));
</code></pre>

blocks|key|835939|text|用Java泛型获取X的列表并返回扩展或实现X的T的列表，甜蜜！|type|unstyled|depth|inlineStyleRanges|entityRanges|data|835940|++++//+the+cast+is+is+actually+checked+via+the+method+API
@SuppressWarnings("unchecked")
public+static+<T+extends+X,+X>+ArrayList<T>+convertToClazz(ArrayList<X>+from,+Class<X>+inClazz,+Class<T>+outClazz)+{
++++ArrayList<T>+to+=+new+ArrayList<T>();
++++for+(X+data+:+from)+{
++++++++to.add((T)+data);
++++}
++++return+to;
}|code-block|syntax|javascript|835941|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|$]]

With Java Generics Takes a list of X and returns a list of T that extends or implements X, Sweet!

<pre><code> // the cast is is actually checked via the method API
@SuppressWarnings("unchecked")
public static &lt;T extends X, X&gt; ArrayList&lt;T&gt; convertToClazz(ArrayList&lt;X&gt; from, Class&lt;X&gt; inClazz, Class&lt;T&gt; outClazz) {
 ArrayList&lt;T&gt; to = new ArrayList&lt;T&gt;();
 for (X data : from) {
 to.add((T) data);
 }
 return to;
}
</code></pre>

blocks|key|1050807|text|一个简单的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1050808|List<Object>+lst++=listOfTypeObject;+++
ArrayList<String>+aryLst+=+new+ArrayList<String>();
++++for+(int+i+=+0;+i+<+lst.size();+i%2B%2B)+{
++++++++aryLst.add(lst.get(i).toString());
++++}|code-block|syntax|javascript|1050809|注意:当列表包含数据类型字符串的所有元素时，这是有效的。|1050810|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|$]]

A simple solution:

<pre><code>List&lt;Object&gt; lst =listOfTypeObject; 
ArrayList&lt;String&gt; aryLst = new ArrayList&lt;String&gt;();
 for (int i = 0; i &lt; lst.size(); i++) {
 aryLst.add(lst.get(i).toString());
 }
</code></pre>

Note: this works when the list contains all the elements of datatype String.

<pre><code>ArrayList&lt;Object&gt; list = new ArrayList&lt;Object&gt;();
list.add(1);
list.add("Java");
list.add(3.14);
System.out.println(list.toString());
</code></pre>

I tried: 

<pre><code>ArrayList&lt;String&gt; list2 = (String)list; 
</code></pre>

But it gave me a compile error.

How can I convert ArrayList<Object> to ArrayList<String>?

Java

ArrayList<Object> list = new ArrayList<Object>();list.add(1);list.add("Java");list.add(3.14);System.out.println(list.toString());我试过了：ArrayList<String> list2 = (String)list; 但它给了我一个编译错误。

问如何将ArrayList<Object>转换为ArrayList<String>？
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