blocks|key|621642|text|String+current+=+elements[i];
if+(i+!=+elements.length+-+1)+{
+++String+next+=+elements[i%2B1];
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|621643|这可以确保你不会得到最后一个元素的ArrayIndexOutOfBoundsException+(那里没有'next‘)。另一个选项是迭代到i+<+elements.length+-+1。这取决于您的要求。|unstyled|offset|length|style|CODE|621644|entityMap^0|0|H|U|1Y|N|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|N|8|@$G|O|H|P|I|J]|$G|Q|H|R|I|J]]|9|@]|A|$]]|$1|K|3|-4|5|F|7|S|8|@]|9|@]|A|$]]]|L|$]]

<pre><code>String current = elements[i];
if (i != elements.length - 1) {
 String next = elements[i+1];
}
</code></pre>

This makes sure you don't get an <code>ArrayIndexOutOfBoundsException</code> for the last element (there is no 'next' there). The other option is to iterate to <code>i &lt; elements.length - 1</code>. It depends on your requirements.

blocks|key|621695|text|String[]+elements+=+{+"a",+"a",+"a",+"a"+};

for(+int+i+=+0;+i+<+elements.length+-+1;+i%2B%2B)
{
++++String+element+=+elements[i];
++++String+nextElement+=+elements[i%2B1];
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|621696|请注意，在本例中，elements.length为4，因此您希望从[0,2]迭代得到元素0,1、1,2和2,3。|unstyled|offset|length|style|CODE|621697|entityMap^0|0|9|F|X|5|18|3|1C|3|1G|3|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|N|8|@$G|O|H|P|I|J]|$G|Q|H|R|I|J]|$G|S|H|T|I|J]|$G|U|H|V|I|J]|$G|W|H|X|I|J]]|9|@]|A|$]]|$1|K|3|-4|5|F|7|Y|8|@]|9|@]|A|$]]]|L|$]]

<pre><code>String[] elements = { "a", "a", "a", "a" };

for( int i = 0; i &lt; elements.length - 1; i++)
{
 String element = elements[i];
 String nextElement = elements[i+1];
}
</code></pre>

Note that in this case, <code>elements.length</code> is 4, so you want to iterate from <code>[0,2]</code> to get elements <code>0,1</code>, <code>1,2</code> and <code>2,3</code>.

blocks|key|621722|text|String[]+elements+=+{+"a",+"a","a","a"+};

for(+int+i=0;+i<elements.length-1;+i%2B%2B)
{
++++String+s1+=+elements[i];
++++String+s2+=+elements[i%2B1];
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|621723|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>String[] elements = { "a", "a","a","a" };

for( int i=0; i&lt;elements.length-1; i++)
{
 String s1 = elements[i];
 String s2 = elements[i+1];
}
</code></pre>

blocks|key|837686|text|你必须保持序列号，就像这样访问array.Use的次数|type|unstyled|depth|inlineStyleRanges|entityRanges|data|837687|int+lookUpTime=0;

++++for(int+i=lookUpTime;i<lookUpTime%2B2+&&+i<elements.length();i%2B%2B)
+++++{
++++//+do+something+with+elements[i]
++++}

lookUpTime%2B%2B;|code-block|syntax|javascript|837688|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 have to maintain the serial how many times you are accessing the array.Use like this

<pre><code>int lookUpTime=0;

 for(int i=lookUpTime;i&lt;lookUpTime+2 &amp;&amp; i&lt;elements.length();i++)
 {
 // do something with elements[i]
 }

lookUpTime++;
</code></pre>

blocks|key|837749|text|我会争辩说，与其测试比elements.length+-+1少的i，不如测试比elements.length少的i+%2B+1。您不是在更改您正在查看的数组的域(即忽略最后一个元素)，而是在每次迭代中更改您正在查看的最大元素。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|837750|String[]+elements+=+{+"a",+"a","a","a"+};

for(int+i+=+0;+i+%2B+1+<+elements.length;+i%2B%2B)+{
++++String+first+=+elements[i];
++++String+second+=+elements[i%2B1];
++++//do+something+with+the+two+strings
}|code-block|syntax|javascript|837751|entityMap^0|B|J|W|1|13|F|1K|5|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|L|$]]

I would argue instead of testing <code>i</code> less than <code>elements.length - 1</code> testing <code>i + 1</code> less than <code>elements.length</code>. You aren't changing the domain of the array that you are looking at (i.e. ignoring the last element), but rather changing the greatest element you are looking at in each iteration.

<pre><code>String[] elements = { "a", "a","a","a" };

for(int i = 0; i + 1 &lt; elements.length; i++) {
 String first = elements[i];
 String second = elements[i+1];
 //do something with the two strings
}
</code></pre>

blocks|key|621960|text|您可以对数组的元素执行enhanced+for+loop+(适用于Java5和更高版本)进行迭代：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|621961|String[]+elements+=+{"a",+"a",+"a",+"a"};+++
for+(String+s:+elements)+{+++++++++++
++++//Do+your+stuff+here
++++System.out.println(s);+
}|code-block|syntax|javascript|621962|entityMap|0|LINK|mutability|MUTABLE|url|https://blogs.oracle.com/CoreJavaTechTips/entry/using_enhanced_for_loops_with^0|B|H|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

You can do an <a href="https://blogs.oracle.com/CoreJavaTechTips/entry/using_enhanced_for_loops_with" rel="noreferrer">enhanced for loop</a> (for java 5 and higher) for iteration on array's elements:

<pre><code>String[] elements = {"a", "a", "a", "a"}; 
for (String s: elements) { 
 //Do your stuff here
 System.out.println(s); 
}
</code></pre>

blocks|key|621969|text|++++String[]+nameArray=+{"John",+"Paul",+"Ringo",+"George"};
++++int+numberOfItems+=+nameArray.length;
++++for+(int+i=0;+i<numberOfItems;+i%2B%2B)
++++{
++++++++String+name+=+nameArray[i];
++++++++System.out.println("Hello+"+%2B+name);
++++}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|621970|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code> String[] nameArray= {"John", "Paul", "Ringo", "George"};
 int numberOfItems = nameArray.length;
 for (int i=0; i&lt;numberOfItems; i++)
 {
 String name = nameArray[i];
 System.out.println("Hello " + name);
 }
</code></pre>

blocks|key|622049|text|如果您正在寻找性能，而迭代的顺序并不相关，则可以使用优化的反向循环进行迭代：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|622050|int+elemLength+=+elements.length;
if(elemLength+<+2){
++//+avoid+ArrayIndexOutOfBoundsException+...
}+else+{
++String+elem1,+elem2;
++for(int+i+=+elemLength+-1;+--i+>=+0;)+{
++++elem1+=+elements[i];
++++elem2+=+elements[i%2B1];
++++//+do+whatever+you+want+with+those+two+strings
++}
}|code-block|syntax|javascript|622051|通过这种方式，您只需检索一次数组的长度，然后递减索引并在一次操作中将其与零进行比较。与零比较是一种非常快速的操作，通常由许多架构优化(比比较数组的长度更容易/更快)。|622052|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

If you are looking for performance and the order of iteration is not relevant, you can iterate using an optimized reverse loop:
<pre><code>int elemLength = elements.length;
if(elemLength &lt; 2){
 // avoid ArrayIndexOutOfBoundsException ...
} else {
 String elem1, elem2;
 for(int i = elemLength -1; --i &gt;= 0;) {
 elem1 = elements[i];
 elem2 = elements[i+1];
 // do whatever you want with those two strings
 }
}
</code></pre>
In such a way you are retrieving the length of the array once, then decrementing the index and comparing with zero in a single operation. Comparing with zero is a very fast operation, often optimized by many architectures (easier / faster than comparing to the length of the array).

I have String array with some components, this array has 5 components and it vary some times. What I would like to do is to iterate through that array and get the first component and the component next to that one. So the first time I would get the component number one and the component number 2, the second time would get the number 2 and 3, the third time number 3 and 4... And so on until you get to the last component. 

This how far I have come: 

<pre><code>String[] elements = { "a", "a","a","a" };

for( int i = 0; i &lt;= elements.length - 1; i++)
{
 // get element number 0 and 1 and put it in a variable, 
 // and the next time get element 1 and 2 and put this in another variable. 
}
</code></pre>

How can I accomplish this?

Iterate through string array in Java

我有一个带有一些组件的字符串数组，这个数组有5个组件，它有时会发生变化。我想要做的是迭代该数组，并获得第一个组件和该组件旁边的组件。所以第一次得到元件号1和元件号2，第二次得到2和3，第三次得到3和4…依此类推，直到您到达最后一个组件。这就是我所走的路：String[] elements = { "a", "a","a...

问在Java中遍历字符串数组
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