blocks|key|1939367|text|这两种说法没有区别。就我个人而言，第二个是首选的。因为您拥有在大括号中指定的所有元素。编译器将帮助您计算数组的大小。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1939368|因此，不需要在赋值操作符之后添加int[]。|offset|length|style|CODE|1939369|entityMap^0|0|G|5|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@$D|L|E|M|F|G]]|9|@]|A|$]]|$1|H|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|I|$]]

There is no difference between the two statements.
Personally speaking, the second one is preferred. Because you have all the elements specified in the braces. The compiler will help you to compute the size of the array. 

So no need to add <code>int[]</code> after the assignment operator.

blocks|key|1725794|text|在本例中，使用第二个，因为它更漂亮，也不那么冗长:)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1725795|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|$]]

In this case, the second one because it's prettier and less verbose :)

blocks|key|1725810|text|正如其他人所提到的，它们是等效的，第二个选项不那么冗长。不幸的是，编译器并不总是能够理解第二种选择：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1725811|public+int[]+getNumbers()+{
+++return+{1,+2,+3};+//illegal+start+of+expression
}|code-block|syntax|javascript|1725812|在这种情况下，您必须使用完整的语法：|1725813|public+int[]+getNumbers()+{
+++return+new+int[]{1,+2,+3};
}|1725814|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|$]]

As others have mentioned, they are equivalent and the second option is less verbose. Unfortunately the compiler isn't always able to understand the second option:

<pre><code>public int[] getNumbers() {
 return {1, 2, 3}; //illegal start of expression
}
</code></pre>

In this case you have to use the full syntax:

<pre><code>public int[] getNumbers() {
 return new int[]{1, 2, 3};
}
</code></pre>

blocks|key|1725830|text|在这种情况下很有用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1725831|void+foo(int[]+array)+{

}|code-block|syntax|javascript|1725832|使用文字调用|1725833|//+works
foo(new+int[]{5,+7})

//illegal
foo({5,+7})|1725834|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|$]]

useful in this situation

<pre><code>void foo(int[] array) {

}
</code></pre>

calling with a literal

<pre><code>// works
foo(new int[]{5, 7})

//illegal
foo({5, 7})
</code></pre>

blocks|key|1939408|text|在你的例子中，这两种风格最终的效果是一样的，都是正确的，第二种更简洁。但实际上这两种风格是不同的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1939409|记住，java中的数组是固定长度的数据结构。创建数组后，必须指定长度。|1939410|如果没有初始化，第一种情况是|1939411|int[]+arr1+=+new+int[5];|code-block|syntax|javascript|1939412|第二种情况是|1939413|int[]+arr2+=+{0,0,0,0,0};|1939414|你看到区别了吗？在这种情况下，首选第一种样式，因为您不必手动键入所有这些默认初始值。|1939415|对我来说，这两种风格之间唯一的最大区别是在没有显式初始化的情况下创建数组。|1939416|entityMap^0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|V|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|W|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|X|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|Y|8|@]|9|@]|A|$]]|$1|M|3|N|5|H|7|Z|8|@]|9|@]|A|$I|J]]|$1|O|3|P|5|6|7|10|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|11|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

In your case, these two styles end up same effect, both correct, with the second one more concise. But actually these two styles are different.

Remember arrays in java are fixed length data structures. Once you create an array, you have to specify the length.

Without initialization, the first case is 

<pre><code>int[] arr1 = new int[5];
</code></pre>

The second case it would be

<pre><code>int[] arr2 = {0,0,0,0,0};
</code></pre>

You see the difference? In this situation, the first style is preferred as you don't have to type all those default initial values manually.

To me, the only big difference between the two styles is when creating an array without explicit initialization.

blocks|key|1725875|text|补充@Paul+Bellora的答案，如果您试图使用三元运算符初始化原始数组，则只有第二种选择有效|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1725876|int[]+arr1+=+false+?+{}+:+{1,2};+//+doesn't+work
int[]+arr2+=+false+?+new+int[]{}+:+new+int[]{1,2};+//+works|code-block|syntax|javascript|1725877|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|$]]

Adding to @Paul Bellora's answer, only the second option will work if you're trying to initialize a primitive array using ternary operator
<pre><code>int[] arr1 = false ? {} : {1,2}; // doesn't work
int[] arr2 = false ? new int[]{} : new int[]{1,2}; // works
</code></pre>

I am wondering,

What's exactly the difference between these two ways of initializing an array of primitives:

<pre><code>int[] arr1 = new int[]{3,2,5,4,1};

int[] arr2 = {3,2,5,4,1};
</code></pre>

and which one is preferred ?

Initialize array of primitives

数据结构

编译器

我在想，这两种初始化原语数组的方法到底有什么不同：int[] arr1 = new int[]{3,2,5,4,1};int[] arr2 = {3,2,5,4,1};哪一个是首选的呢？

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