blocks|key|1043367|text|你不能(没有不安全的类型转换)。您只能从它们中读取。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1043368|问题是你不知道这个列表到底是什么。它可以是Number的任何子类的列表，所以当您尝试将一个元素放入其中时，您不知道该元素实际上是否适合该列表。|1043369|例如，列表可能是Byte的列表，因此将Float放入其中将是错误的。|offset|length|style|CODE|1043370|entityMap^0|0|0|8|4|J|5|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|M|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|N|8|@$F|O|G|P|H|I]|$F|Q|G|R|H|I]]|9|@]|A|$]]|$1|J|3|-4|5|6|7|S|8|@]|9|@]|A|$]]]|K|$]]

You can't (without unsafe casts). You can only read from them.

The problem is that you don't know what exactly the list is a list of. It could be a list of any subclass of Number, so when you try to put an element into it, you don't know that the element actually fits into the list.

For example the List might be a list of <code>Byte</code>s, so it would be an error to put a <code>Float</code> into it.

blocks|key|1258329|text|你可以这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1258330|++List<Number>+foo3+=+new+ArrayList<Number>();++++++
++foo3.add(3);|code-block|syntax|javascript|1258331|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 could do this instead:

<pre><code> List&lt;Number&gt; foo3 = new ArrayList&lt;Number&gt;(); 
 foo3.add(3);
</code></pre>

blocks|key|1258405|text|您可以通过创建一个具有不同类型的列表引用来对其进行虚构。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1258406|(这些是sepp2k提到的“不安全类型转换”。)|1258407|List<?+extends+Number>+list+=+new+ArrayList<Integer>();

//+This+will+not+compile
//list.add(100);

//+WORKS,+BUT+NOT+IDEAL
List+untypedList+=+(List)list;
//+It+will+let+you+add+a+number
untypedList.add(200);
//+But+it+will+also+let+you+add+a+String!++BAD!
untypedList.add("foo");

//+YOU+PROBABLY+WANT+THIS
//+This+is+safer,+because+it+will+(partially)+check+the+type+of+anything+you+add
List<Number>+superclassedList+=+(List<Number>)(List<?>)list;
//+It+will+let+you+add+an+integer
superclassedList.add(200);
//+It+won't+let+you+add+a+String
//superclassedList.add("foo");
//+But+it+will+let+you+add+a+Float,+which+isn't+really+correct
superclassedList.add(3.141);
//+********************
//+So+you+are+responsible+for+ensuring+you+only+add/set+Integers+when+you+have
//+been+given+an+ArrayList<Integer>
//+********************

//+EVEN+BETTER
//+If+you+can,+if+you+know+the+type,+then+use+List<Integer>+instead+of+List<Number>
List<Integer>+trulyclassedList+=+(List<Integer>)(List<?>)list;
//+That+will+prevent+you+from+adding+a+Float
//trulyclassedList.add(3.141);

System.out.println("list:+"+%2B+list);|code-block|syntax|javascript|1258408|因为untypedList、superclassedList和trulyclassedList只是对list的引用，所以仍然需要向原始ArrayList添加元素。|offset|length|style|CODE|1258409|在上面的示例中，您实际上不需要使用(List<?>)，但是您可能需要在代码中使用它，这取决于提供给您的list的类型。|1258410|注意，使用?会给你编译器警告，直到你把它放在你的函数上面：|1258411|@SuppressWarnings("unchecked")|1258412|entityMap^0|0|0|0|2|B|E|G|V|G|1E|4|0|H|9|1F|4|0|5|1|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|X|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|Z|8|@$K|10|L|11|M|N]|$K|12|L|13|M|N]|$K|14|L|15|M|N]|$K|16|L|17|M|N]]|9|@]|A|$]]|$1|O|3|P|5|6|7|18|8|@$K|19|L|1A|M|N]|$K|1B|L|1C|M|N]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1D|8|@$K|1E|L|1F|M|N]]|9|@]|A|$]]|$1|S|3|T|5|F|7|1G|8|@]|9|@]|A|$G|H]]|$1|U|3|-4|5|6|7|1H|8|@]|9|@]|A|$]]]|V|$]]

You can fudge it by creating a reference to the List with a different type.
(These are the &quot;unsafe casts&quot; mentioned by sepp2k.)
<pre class="lang-java prettyprint-override"><code>List&lt;? extends Number&gt; list = new ArrayList&lt;Integer&gt;();

// This will not compile
//list.add(100);

// WORKS, BUT NOT IDEAL
List untypedList = (List)list;
// It will let you add a number
untypedList.add(200);
// But it will also let you add a String! BAD!
untypedList.add(&quot;foo&quot;);

// YOU PROBABLY WANT THIS
// This is safer, because it will (partially) check the type of anything you add
List&lt;Number&gt; superclassedList = (List&lt;Number&gt;)(List&lt;?&gt;)list;
// It will let you add an integer
superclassedList.add(200);
// It won't let you add a String
//superclassedList.add(&quot;foo&quot;);
// But it will let you add a Float, which isn't really correct
superclassedList.add(3.141);
// ********************
// So you are responsible for ensuring you only add/set Integers when you have
// been given an ArrayList&lt;Integer&gt;
// ********************

// EVEN BETTER
// If you can, if you know the type, then use List&lt;Integer&gt; instead of List&lt;Number&gt;
List&lt;Integer&gt; trulyclassedList = (List&lt;Integer&gt;)(List&lt;?&gt;)list;
// That will prevent you from adding a Float
//trulyclassedList.add(3.141);

System.out.println(&quot;list: &quot; + list);
</code></pre>
Because <code>untypedList</code>, <code>superclassedList</code> and <code>trulyclassedList</code> are just references to <code>list</code>, you will still be adding elements to the original ArrayList.
You don't actually need to use <code>(List&lt;?&gt;)</code> in the example above, but you might need it in your code, depending on the type of <code>list</code> you were given.
Note that using <code>?</code> will give you compiler warnings, until you put this above your function:
<pre class="lang-java prettyprint-override"><code>@SuppressWarnings(&quot;unchecked&quot;)
</code></pre>

blocks|key|1258455|text|因为3是没有扩展Number的primitive+(int)，所以您可以添加它的装箱类型(Integer.valueOf(3))。尽管它在IDEA中显示为错误，但它仍然可以正常执行。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1258456|entityMap^0|8|6|F|F|18|K|1X|4|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]|$9|O|A|P|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|G|$]]

Because 3 is <code>primitive (int)</code>, which does not extend <code>Number</code>, you can add its boxed type <code>(Integer.valueOf(3))</code>. Although it is displayed as an error in <code>IDEA</code>, it can still execute normally.

I have a List which is declared like this :

<pre><code> List&lt;? extends Number&gt; foo3 = new ArrayList&lt;Integer&gt;();
</code></pre>

I tried to add 3 to foo3. However I get an error message like this:

<pre><code>The method add(capture#1-of ? extends Number) in the type List&lt;capture#1-of ?
extends Number&gt; is not applicable for the arguments (ExtendsNumber)
</code></pre>

How can I add to List<? extends Number> data structures?

编译器

我有一个如下声明的列表： List<? extends Number> foo3 = new ArrayList<Integer>();我尝试将3加到foo3中。但是，我得到如下的错误消息：The method add(capture#1-of ? extends Number) in the type List<ca...

问如何添加到List<？扩展Number>数据结构？
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