blocks|key|4421867|text|通用信息在运行时是erased。JVM不知道您的列表是List<String>还是List<Integer>+(在运行时，List<T>中的T被解析为Object)，因此惟一可能的数组类型是Object[]。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4421868|您可以使用toArray(T[]+array)+-在这种情况下，JVM可以使用给定数组的类，您可以在ArrayList实现中看到它：|4421869|public+<T>+T[]+toArray(T[]+a)+{
++++if+(a.length+<+size)
++++++++//+Make+a+new+array+of+a's+runtime+type,+but+my+contents:
++++++++return+(T[])+Arrays.copyOf(elementData,+size,+a.getClass());|code-block|syntax|javascript|4421870|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.oracle.com/javase/tutorial/java/generics/genTypes.html^0|R|C|15|D|1P|7|1Y|1|23|6|2N|8|9|6|0|0|5|I|1E|9|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$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]|$9|15|A|16|B|C]]|D|@$9|17|A|18|1|19]]|E|$]]|$1|F|3|G|5|6|7|1A|8|@$9|1B|A|1C|B|C]|$9|1D|A|1E|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|1F|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|1G|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

Generic information is <a href="https://docs.oracle.com/javase/tutorial/java/generics/genTypes.html" rel="noreferrer">erased</a> at runtime. JVM does not know whether your list is <code>List&lt;String&gt;</code> or <code>List&lt;Integer&gt;</code> (at runtime <code>T</code> in <code>List&lt;T&gt;</code> is resolved as <code>Object</code>), so the only possible array type is <code>Object[]</code>.

You can use <code>toArray(T[] array)</code> though - in this case JVM can use the class of a given array, you can see it in the <code>ArrayList</code> implementation:

<pre><code>public &lt;T&gt; T[] toArray(T[] a) {
 if (a.length &lt; size)
 // Make a new array of a's runtime type, but my contents:
 return (T[]) Arrays.copyOf(elementData, size, a.getClass());
</code></pre>

blocks|key|4421787|text|如果查看Javadoc+for+the+List+interface，您会注意到toArray的第二种形式：<T>+T[]+toArray(T[]+a)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4421788|事实上，Javadoc甚至给出了一个示例，说明如何准确地完成您想要做的事情：|4421789|String[]+y+=+x.toArray(new+String[0]);|4421790|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.oracle.com/javase/7/docs/api/java/util/List.html^0|K|4|14|7|1I|M|4|U|0|0|0|0|12|0^^$0|@$1|2|3|4|5|6|7|R|8|@$9|S|A|T|B|C]|$9|U|A|V|B|C]|$9|W|A|X|B|C]]|D|@$9|Y|A|Z|1|10]]|E|$]]|$1|F|3|G|5|6|7|11|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|12|8|@$9|13|A|14|B|C]]|D|@]|E|$]]|$1|J|3|-4|5|6|7|15|8|@]|D|@]|E|$]]]|K|$L|$5|M|N|O|E|$P|Q]]]]

If you look at the <a href="https://docs.oracle.com/javase/7/docs/api/java/util/List.html" rel="noreferrer">Javadoc for the <code>List</code> interface</a>, you'll notice a second form of <code>toArray</code>: <code>&lt;T&gt; T[] toArray(T[] a)</code>.

In fact, the Javadoc even gives an example of how to do exactly what you want to do:

<code>String[] y = x.toArray(new String[0]);</code>

blocks|key|4421814|text|需要注意的是，Java中的数组在运行时知道它们的组件类型。String[]和Integer[]在运行时是不同的类，您可以在运行时向数组询问它们的组件类型。因此，需要在运行时使用组件类型(通过在编译时使用new+String[...]对可重用的组件类型进行硬编码，或者使用Array.newInstance()并传递类对象)来创建数组。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4421815|另一方面，泛型中的类型参数在运行时不存在。在运行时，ArrayList<String>和ArrayList<Integer>之间完全没有区别。这一切都只是ArrayList而已。|4421816|这就是为什么您不能只获取List<String>并获取String[]而不以某种方式单独传入组件类型的根本原因--您必须从没有组件类型信息的对象中获取组件类型信息。显然，这是不可能的。|4421817|entityMap^0|T|8|12|9|2T|F|3R|J|0|Q|H|18|I|25|9|0|C|C|R|8|0^^$0|@$1|2|3|4|5|6|7|L|8|@$9|M|A|N|B|C]|$9|O|A|P|B|C]|$9|Q|A|R|B|C]|$9|S|A|T|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|11|8|@$9|12|A|13|B|C]|$9|14|A|15|B|C]]|D|@]|E|$]]|$1|J|3|-4|5|6|7|16|8|@]|D|@]|E|$]]]|K|$]]

The pertinent thing to note is that arrays in Java know their component type at runtime. <code>String[]</code> and <code>Integer[]</code> are different classes at runtime, and you can ask arrays for their component type at runtime. Therefore, a component type is needed at runtime (either by hard-coding a reifiable component type at compile time with <code>new String[...]</code>, or using <code>Array.newInstance()</code> and passing a class object) to create an array.

On the other hand, type arguments in generics do not exist at runtime. There is absolutely no difference at runtime between an <code>ArrayList&lt;String&gt;</code> and a <code>ArrayList&lt;Integer&gt;</code>. It is all just <code>ArrayList</code>.

That's the fundamental reason why you can't just take a <code>List&lt;String&gt;</code> and get a <code>String[]</code> without passing in the component type separately somehow -- you would have to get component type information out of something that doesn't have component type information. Clearly, this is impossible.

blocks|key|4421970|text|您首先要了解的是，ArrayList拥有的只是一个Object数组|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4421971|+++transient+Object[]+elementData;|code-block|syntax|javascript|4421972|当谈到T[]失败的原因时，它是因为你不能在没有Class<T>的情况下获得泛型类型的数组，这是因为java的类型擦除(+there+is+a+more+explanation和how+to+create+one)。堆上的array[]动态地知道它的类型，您不能将int[]转换为String[]。同样的原因，您不能将Object[]转换为T[]。|4421973|+++int[]+ints+=+new+int[3];
+++String[]+strings+=+(String[])+ints;//java:+incompatible+types:+int[]+cannot+be+converted+to+java.lang.String[]

+++public+<T>+T[]+a()+{
++++++Object[]+objects+=+new+Object[3];
++++++return+(T[])objects;
+++}
+++//ClassCastException:+[Ljava.lang.Object;+cannot+be+cast+to+[Ljava.lang.Integer;
+++Integer[]+a+=+new+LearnArray().<Integer>a();|4421974|但是您放到array中的只是一个类型为E(由编译器检查)的对象，因此您可以将其强制转换为E，这是安全和正确的。|4421975|++return+(E)+elementData[index];|4421976|简而言之，你不能通过cast获得没有的东西。您只有Object[]，所以toArray()可以只返回Object[](否则，您必须给它一个Class<T>才能用这个类型创建一个新的数组)。你把E放在ArrayList<E>里，你可以用get()得到一个E。|4421977|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/2927391/whats-the-reason-i-cant-create-generic-array-types-in-java|1|https://stackoverflow.com/questions/529085/how-to-create-a-generic-array-in-java^0|9|9|P|6|0|0|3|3|N|8|32|7|3N|5|3V|8|4E|8|4P|3|1O|R|0|2G|H|1|0|0|5|5|J|1|18|1|0|0|P|8|10|9|1E|8|1X|8|2O|1|2R|C|39|5|3I|1|0^^$0|@$1|2|3|4|5|6|7|14|8|@$9|15|A|16|B|C]|$9|17|A|18|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|19|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1A|8|@$9|1B|A|1C|B|C]|$9|1D|A|1E|B|C]|$9|1F|A|1G|B|C]|$9|1H|A|1I|B|C]|$9|1J|A|1K|B|C]|$9|1L|A|1M|B|C]|$9|1N|A|1O|B|C]]|D|@$9|1P|A|1Q|1|1R]|$9|1S|A|1T|1|1U]]|E|$]]|$1|M|3|N|5|H|7|1V|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|1W|8|@$9|1X|A|1Y|B|C]|$9|1Z|A|20|B|C]|$9|21|A|22|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|H|7|23|8|@]|D|@]|E|$I|J]]|$1|S|3|T|5|6|7|24|8|@$9|25|A|26|B|C]|$9|27|A|28|B|C]|$9|29|A|2A|B|C]|$9|2B|A|2C|B|C]|$9|2D|A|2E|B|C]|$9|2F|A|2G|B|C]|$9|2H|A|2I|B|C]|$9|2J|A|2K|B|C]]|D|@]|E|$]]|$1|U|3|-4|5|6|7|2L|8|@]|D|@]|E|$]]]|V|$W|$5|X|Y|Z|E|$10|11]]|12|$5|X|Y|Z|E|$10|13]]]]

The very first thing you have to understand is what <code>ArrayList</code> own is just an array of <code>Object</code>

<pre><code> transient Object[] elementData;
</code></pre>

When it comes to the reason why <code>T[]</code> is fail, it because you can't get an array of generic type without a <code>Class&lt;T&gt;</code> and this is because java's type erase( <a href="https://stackoverflow.com/questions/2927391/whats-the-reason-i-cant-create-generic-array-types-in-java">there is a more explanation</a> and <a href="https://stackoverflow.com/questions/529085/how-to-create-a-generic-array-in-java">how to create one</a>). And the <code>array[]</code> on the heap knows its type dynamically and you can't cast <code>int[]</code> to <code>String[]</code>. The same reason, you can't cast <code>Object[]</code> to <code>T[]</code>.

<pre><code> int[] ints = new int[3];
 String[] strings = (String[]) ints;//java: incompatible types: int[] cannot be converted to java.lang.String[]

 public &lt;T&gt; T[] a() {
 Object[] objects = new Object[3];
 return (T[])objects;
 }
 //ClassCastException: [Ljava.lang.Object; cannot be cast to [Ljava.lang.Integer;
 Integer[] a = new LearnArray().&lt;Integer&gt;a();
</code></pre>

But what you put into the <code>array</code> is just a object which type is <code>E</code>(which is checked by compiler), so you can just cast it to <code>E</code> which is safe and correct.

<pre><code> return (E) elementData[index];
</code></pre>

In short, you can't get what don't have by cast. You have just <code>Object[]</code>, so <code>toArray()</code> can just return <code>Object[]</code>(otherwise, you have to give it a <code>Class&lt;T&gt;</code> to make a new array with this type). You put <code>E</code> in <code>ArrayList&lt;E&gt;</code>, you can get a <code>E</code> with <code>get()</code>.

blocks|key|4421897|text|可以创建给定(已知)类型的“通用”数组。通常我会在我的代码中使用类似这样的东西。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4421898|public+static+<T>+T[]+toArray(Class<T>+type,+ArrayList<T>+arrList)+{
++++if+((arrList+==+null)+%7C%7C+(arrList.size()+==+0))+return+null;
++++Object+arr+=+Array.newInstance(type,+arrList.size());
++++for+(int+i=0;+i+<+arrList.size();+i%2B%2B)+Array.set(arr,+i,+arrList.get(i));
++++return+(T[])arr;
}|code-block|syntax|javascript|4421899|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|$]]

It is possible to create a "generic" array of the given(known) type. Normally I use something like this in my code.

<pre><code>public static &lt;T&gt; T[] toArray(Class&lt;T&gt; type, ArrayList&lt;T&gt; arrList) {
 if ((arrList == null) || (arrList.size() == 0)) return null;
 Object arr = Array.newInstance(type, arrList.size());
 for (int i=0; i &lt; arrList.size(); i++) Array.set(arr, i, arrList.get(i));
 return (T[])arr;
}
</code></pre>

Say you have an arraylist defined as follows:

<pre><code>ArrayList&lt;String&gt; someData = new ArrayList&lt;&gt;();
</code></pre>

Later on in your code, because of generics you can say this:

<pre><code>String someLine = someData.get(0);
</code></pre>

And the compiler knows outright that it will be getting a string. Yay generics! However, this will fail:

<pre><code>String[] arrayOfData = someData.toArray();
</code></pre>

<code>toArray()</code> will always return an array of Objects, not of the generic that was defined. Why does the <code>get(x)</code> method know what it is returning, but <code>toArray()</code> defaults to Objects?

Java generics in ArrayList.toArray()

假设您有一个如下定义的arraylist：ArrayList<String> someData = new ArrayList<>();在稍后的代码中，由于泛型，您可以这样说：String someLine = someData.get(0);编译器直接知道它将得到一个字符串。耶，泛型！但是，这将失败：String[]...

问ArrayList.toArray()中的Java泛型
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