blocks|key|1087213|text|不，你必须遍历它们。抱歉的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1087214|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|$]]

Nope; you've got to iterate over them. Sorry.

blocks|key|1087238|text|public+int[]+toInt(Set<Integer>+set)+{
++int[]+a+=+new+int[set.size()];
++int+i+=+0;
++for+(Integer+val+:+set)+a[i%2B%2B]+=+val;
++return+a;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1087239|既然我已经为你写好了代码，它就不再是手册了，不是吗？;)|unstyled|1087240|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>public int[] toInt(Set&lt;Integer&gt; set) {
 int[] a = new int[set.size()];
 int i = 0;
 for (Integer val : set) a[i++] = val;
 return a;
}
</code></pre>

Now that I wrote the code for you it's not that manual anymore, is it? ;)

blocks|key|1302417|text|阿帕奇的ArrayUtils有这个(它仍然迭代behind+the+scenes)：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1302418|doSomething(ArrayUtils.toPrimitive(hashset.toArray()));|code-block|syntax|javascript|1302419|它们总是检查这类事情的好地方。|1302420|entityMap|0|LINK|mutability|MUTABLE|url|http://commons.apache.org/proper/commons-lang/javadocs/api-release/org/apache/commons/lang3/ArrayUtils.html|1|https://github.com/apache/commons-lang/blob/LANG_3_2_RC1/src/main/java/org/apache/commons/lang3/ArrayUtils.java^0|4|A|0|N|H|1|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]|$A|Y|B|Z|1|10]]|C|$]]|$1|D|3|E|5|F|7|11|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|12|8|@]|9|@]|C|$]]|$1|K|3|-4|5|6|7|13|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]|S|$5|N|O|P|C|$Q|T]]]]

Apache's <a href="http://commons.apache.org/proper/commons-lang/javadocs/api-release/org/apache/commons/lang3/ArrayUtils.html" rel="nofollow noreferrer">ArrayUtils</a> has this (it still iterates <a href="https://github.com/apache/commons-lang/blob/LANG_3_2_RC1/src/main/java/org/apache/commons/lang3/ArrayUtils.java" rel="nofollow noreferrer">behind the scenes</a>):
<pre><code>doSomething(ArrayUtils.toPrimitive(hashset.toArray()));
</code></pre>
They're always a good place to check for things like this.

blocks|key|1302462|text|您还可以使用toArray()方法的toArray(T[]+contents)变体。创建一个与HashSet大小相同的int空数组，然后将其传递给toArray()方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1302463|Integer[]+myarray+=+new+Integer[hashset.size()];
doSomething(hashset.toArray(myarray));|code-block|syntax|javascript|1302464|您必须更改doSomething()函数以接受Integer[]数组，而不是int[]。如果这样做不可行，则需要将toArray返回的值数组转换为int[]。|offset|length|style|CODE|1302465|entityMap^0|0|0|5|D|N|9|12|5|1L|7|21|5|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]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]|$I|T|J|U|K|L]|$I|V|J|W|K|L]|$I|X|J|Y|K|L]|$I|Z|J|10|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|N|$]]

You could also use the toArray(T[] contents) variant of the toArray() method. Create an empty array of ints of the same size as the HashSet, and then pass it to the toArray() method:

<pre><code>Integer[] myarray = new Integer[hashset.size()];
doSomething(hashset.toArray(myarray));
</code></pre>

You'd have to change the <code>doSomething()</code> function to accept an <code>Integer[]</code> array instead of <code>int[]</code>. If that is not feasible, you'd have convert the array of values returned by <code>toArray</code> to <code>int[]</code>.

blocks|key|1302492|text|即使没有Apache+Utils，您也可以将Set<Integer>转换为Integer[]：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1302493|Set<Integer>+myset+=+new+HashSet<Integer>();
Integer[]+array+=+myset.toArray(new+Integer[0]);|code-block|syntax|javascript|1302494|但是，如果您需要int[]，则必须遍历集合。|1302495|entityMap^0|M|C|11|9|0|0|8|5|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|N|$]]

You can convert a <code>Set&lt;Integer&gt;</code> to <code>Integer[]</code> even without Apache Utils:

<pre><code>Set&lt;Integer&gt; myset = new HashSet&lt;Integer&gt;();
Integer[] array = myset.toArray(new Integer[0]);
</code></pre>

However, if you need <code>int[]</code> you have to iterate over the set.

blocks|key|1087352|text|您可以只使用Guava：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1087353|Ints.toArray(Collection<?+extends+Number>+collection)|code-block|syntax|javascript|1087354|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 just use Guava's:

<pre><code>Ints.toArray(Collection&lt;? extends Number&gt; collection)
</code></pre>

blocks|key|1087407|text|您可以使用Java8+streams从任何Collection<Integer>+(包括HashSet<Integer>)创建int[]：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1087408|int[]+array+=+coll.stream().mapToInt(Number::intValue).toArray();|code-block|syntax|javascript|1087409|当然，该库仍在代表您迭代集合(或其他流源)。|1087410|除了简洁和没有外部库依赖之外，如果你有一个非常大的集合要复制，streams还可以让你并行。|1087411|entityMap^0|L|J|18|G|1R|5|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@]|D|@]|E|$]]|$1|M|3|N|5|6|7|Z|8|@]|D|@]|E|$]]|$1|O|3|-4|5|6|7|10|8|@]|D|@]|E|$]]]|P|$]]

You can create an <code>int[]</code> from any <code>Collection&lt;Integer&gt;</code> (including a <code>HashSet&lt;Integer&gt;</code>) using Java 8 streams:

<pre><code>int[] array = coll.stream().mapToInt(Number::intValue).toArray();
</code></pre>

The library is still iterating over the collection (or other stream source) on your behalf, of course.

In addition to being concise and having no external library dependencies, streams also let you go parallel if you have a really big collection to copy.

blocks|key|1087479|text|尝尝这个。使用java+8。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1087480|++++Set<Integer>+set+=+new+HashSet<>();
++++set.add(43);
++++set.add(423);
++++set.add(11);
++++set.add(44);
++++set.add(56);
++++set.add(422);
++++set.add(34);
++++int[]+arr+=+set.stream().mapToInt(Integer::intValue).toArray();|code-block|syntax|javascript|1087481|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|$]]

Try this.
Using java 8.

<pre><code> Set&lt;Integer&gt; set = new HashSet&lt;&gt;();
 set.add(43);
 set.add(423);
 set.add(11);
 set.add(44);
 set.add(56);
 set.add(422);
 set.add(34);
 int[] arr = set.stream().mapToInt(Integer::intValue).toArray();
</code></pre>

I've got a <code>HashSet&lt;Integer&gt;</code> with a bunch of <code>Integers</code> in it. I want to turn it into an array, but calling

<pre><code>hashset.toArray();
</code></pre>

returns an <code>Object[]</code>. Is there a better way to cast it to an array of <code>int</code> other than iterating through every element manually? I want to pass the array to

<pre><code>void doSomething(int[] arr)
</code></pre>

which won't accept the Object[] array, even if I try casting it like 

<pre><code>doSomething((int[]) hashSet.toArray());
</code></pre>

How can I convert a Java HashSet<Integer> to a primitive int array?

我有一个里面有一堆Integers的HashSet<Integer>。我想把它变成一个数组，但是调用hashset.toArray();返回Object[]。除了手动遍历每个元素之外，还有更好的方法将其转换为int数组吗？我想将数组传递给void doSomething(int[] arr)它不会接受Object[]数...

问如何将Java数组转换为原始的HashSet<Integer>数组？
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