blocks|key|4421936|text|Arrays.fill()将用相同的值填充现有数组。原语和Objects有不同的变体。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4421937|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/6/docs/api/java/util/Arrays.html#fill%2528boolean%255B%255D,%2520boolean%2529^0|T|7|0|D|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@$9|O|A|P|B|C]]|D|@$9|Q|A|R|1|S]]|E|$]]|$1|F|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|G|$H|$5|I|J|K|E|$L|M]]]]

<a href="http://docs.oracle.com/javase/6/docs/api/java/util/Arrays.html#fill%28boolean%5B%5D,%20boolean%29" rel="noreferrer">Arrays.fill()</a> will fill an existing array with the same value. Variants exist for primitives and <code>Objects</code>.

blocks|key|4904589|text|您可以使用以下命令进行尝试：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4904590|boolean[]+array+=+new+boolean[5];
Arrays.fill(array,+false);|code-block|syntax|javascript|4904591|使用手动数组填充的第二种方法：|4904592|boolean[]+array+=+new+boolean[]+{false,+false,+false,+false,+false};|4904593|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]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

You can try it with:

<pre><code>boolean[] array = new boolean[5];
Arrays.fill(array, false);
</code></pre>

Second method with manual array fill:

<pre><code>boolean[] array = new boolean[] {false, false, false, false, false};
</code></pre>

blocks|key|4904600|text|对于该特定示例，默认情况下，nothing，a+boolean[]将被初始化为[false,+false,+...]。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4904601|如果你想用非默认值初始化你的数组，你需要循环或者使用Arrays.fill，它会为你做循环。|4904602|entityMap^0|O|9|13|J|0|Q|B|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]|$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|R|8|@]|D|@]|E|$]]]|I|$]]

For that specific example, nothing, a <code>boolean[]</code> will be initialised to <code>[false, false, ...]</code> by default.

If you want to initialise your array with non-default values, you will need to loop or use <code>Arrays.fill</code> which does the loop for you.

blocks|key|1071921|text|Arrays.fill(...)就是你要找的东西。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1071922|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/7/docs/api/java/util/Arrays.html#fill%2528boolean%255B%255D,%2520boolean%2529^0|0|G|0|G|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@$9|O|A|P|B|C]]|D|@$9|Q|A|R|1|S]]|E|$]]|$1|F|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|G|$H|$5|I|J|K|E|$L|M]]]]

<a href="http://docs.oracle.com/javase/7/docs/api/java/util/Arrays.html#fill%28boolean%5B%5D,%20boolean%29" rel="nofollow"><code>Arrays.fill(...)</code></a> is what you are looking for.

blocks|key|4904606|text|试试这个..。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4904607|+Boolean+[]+data+=+new+Boolean[20];
++Arrays.fill(data,new+Boolean(false));|code-block|syntax|javascript|4904608|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 .. 

<pre><code> Boolean [] data = new Boolean[20];
 Arrays.fill(data,new Boolean(false));
</code></pre>

blocks|key|4904636|text|List<Integer>+copies+=+Collections.nCopies(copiesCount,+value);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4904637|javadoc+here。|unstyled|offset|length|4904638|这比'Arrays.fill‘解决方案更好的原因有几个：|4904639|4904640|它很好很流畅，|ordered-list-item|4904641|它消耗更少的内存(参见source+code)，这对于复制大量的副本或巨大的对象是很重要的，|4904642|它创建一个不可变的列表，|4904643|它可以创建一个非原语类型的对象的副本列表。但是应该谨慎使用，因为元素本身不会被复制，get()方法将为每个索引返回相同的值。最好是提供一个不可变的对象进行复制，或者确保它不会被更改。|4904644|4904645|列表比数组更酷:)但是如果你真的-真的-真的想要一个数组，那么你可以这样做：|4904646|Integer[]+copies+=+Collections.nCopies(copiesCount,+value)
++++++++++++++++++++++++++++++.toArray(new+Integer[copiesCount]);|4904647|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/7/docs/api/java/util/Collections.html#nCopies%2528int,%2520T%2529|1|http://grepcode.com/file/repository.grepcode.com/java/root/jdk/openjdk/7-b147/java/util/Collections.java#3463^0|0|8|4|0|0|0|0|0|B|B|1|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|19|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|1A|8|@]|9|@$G|1B|H|1C|1|1D]]|A|$]]|$1|I|3|J|5|F|7|1E|8|@]|9|@]|A|$]]|$1|K|3|-4|5|F|7|1F|8|@]|9|@]|A|$]]|$1|L|3|M|5|N|7|1G|8|@]|9|@]|A|$]]|$1|O|3|P|5|N|7|1H|8|@]|9|@$G|1I|H|1J|1|1K]]|A|$]]|$1|Q|3|R|5|N|7|1L|8|@]|9|@]|A|$]]|$1|S|3|T|5|N|7|1M|8|@]|9|@]|A|$]]|$1|U|3|-4|5|F|7|1N|8|@]|9|@]|A|$]]|$1|V|3|W|5|F|7|1O|8|@]|9|@]|A|$]]|$1|X|3|Y|5|6|7|1P|8|@]|9|@]|A|$B|C]]|$1|Z|3|-4|5|F|7|1Q|8|@]|9|@]|A|$]]]|10|$11|$5|12|13|14|A|$15|16]]|17|$5|12|13|14|A|$15|18]]]]

<pre><code>List&lt;Integer&gt; copies = Collections.nCopies(copiesCount, value);
</code></pre>

javadoc <a href="http://docs.oracle.com/javase/7/docs/api/java/util/Collections.html#nCopies%28int,%20T%29" rel="noreferrer">here</a>.

This is better than the 'Arrays.fill' solution by several reasons:

<ol>
<li>it's nice and smooth,</li>
<li>it consumes less memory (see <a href="http://grepcode.com/file/repository.grepcode.com/java/root/jdk/openjdk/7-b147/java/util/Collections.java#3463" rel="noreferrer">source code</a>) which is significant for a huge copies amount or huge objects to copy,</li>
<li>it creates an immutable list,</li>
<li>it can create a list of copies of an object of a non-primitive type. That should be used with caution though because the element itself will not be duplicated and get() method will return the same value for every index. It's better to provide an immutable object for copying or make sure it's not going to be changed.</li>
</ol>

And lists are cooler than arrays :) But if you really-really-really want an array – then you can do the following:

<pre><code>Integer[] copies = Collections.nCopies(copiesCount, value)
 .toArray(new Integer[copiesCount]);
</code></pre>

blocks|key|4904654|text|或者你也可以用低级的方式来做。创建一个包含n个元素的数组，并遍历放入相同元素的所有元素。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4904655|int[]+array+=+new+int[n];

for+(int+i+=+0;+i+<+n;+i%2B%2B)
{
++++array[i]+=+5;
}|code-block|syntax|javascript|4904656|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|$]]

Or you can do it in the low level way. Make an array with n elements and iterate through all the element where the same element is put in.

<pre><code>int[] array = new int[n];

for (int i = 0; i &lt; n; i++)
{
 array[i] = 5;
}
</code></pre>

I want to create an array of size <code>n</code> with the same value at every index in the array. What's the best way to do this in Java?

For example, if <code>n</code> is 5 and the value is the boolean <code>false</code>, the array should be:

<pre><code>= [false, false, false, false, false]
</code></pre>

Create an array with n copies of the same value/object?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我想创建一个大小为n的数组，数组中的每个索引都有相同的值。在Java中做这件事的最好方法是什么？例如，如果n为5，值为布尔型false，则数组应为：= [false, false, false, false, false]

问是否创建具有相同值/对象的n个副本的数组？
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问是否创建具有相同值/对象的n个副本的数组？EN