blocks|key|608452|text|你不需要重新发明轮子，here的toArray()方法：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|608453|String+[]dsf+=+new+String[al.size()];
al.toArray(dsf);|code-block|syntax|javascript|608454|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/6/docs/api/java/util/ArrayList.html#toArray%2528%2529^0|G|9|B|4|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@$9|V|A|W|1|X]]|E|$]]|$1|F|3|G|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

You don't need to reinvent the wheel, <a href="http://docs.oracle.com/javase/6/docs/api/java/util/ArrayList.html#toArray%28%29">here</a>'s the <code>toArray()</code> method:

<pre><code>String []dsf = new String[al.size()];
al.toArray(dsf);
</code></pre>

blocks|key|392417|text|List<String>+list=new+ArrayList<String>();
list.add("sravan");
list.add("vasu");
list.add("raki");
String+names[]=list.toArray(new+String[list.size()])|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|392418|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>List&lt;String&gt; list=new ArrayList&lt;String&gt;();
list.add("sravan");
list.add("vasu");
list.add("raki");
String names[]=list.toArray(new String[list.size()])
</code></pre>

blocks|key|608703|text|List<String>+list=new+ArrayList<String>();
list.add("sravan");
list.add("vasu");
list.add("raki");+
String+names[]=list.toArray(new+String[0]);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|608704|如果你看到最后一行(新的字符串)，你不必给出大小，有时我们不知道列表的长度，所以从给它0开始，构造的数组将调整大小。|unstyled|608705|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>List&lt;String&gt; list=new ArrayList&lt;String&gt;();
list.add("sravan");
list.add("vasu");
list.add("raki"); 
String names[]=list.toArray(new String[0]);
</code></pre>

if you see the last line (new String[0]), you don't have to give the size, there are time when we don't know the length of the list, so to start with giving it as 0 , the constructed array will resize.

blocks|key|608628|text|String[]+values+=+new+String[arrayList.size()];
++++++++for+(int+i+=+0;+i+<+arrayList.size();+i%2B%2B)+{
++++++++++++values[i]+=+arrayList.get(i).type;
++++++++}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|608629|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[] values = new String[arrayList.size()];
 for (int i = 0; i &lt; arrayList.size(); i++) {
 values[i] = arrayList.get(i).type;
 }
</code></pre>

blocks|key|392348|text|根据我对这个问题的理解，你对迭代所做的事情并不是错误的。它为您提供了一个有效的String对象数组。但是，就像在另一个答案中提到的那样，使用可用于ArrayList+object+=>+http://docs.oracle.com/javase/1.5.0/docs/api/java/util/ArrayList.html#toArray%2528%2529的toArray()方法更容易|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|392349|这只是个附注。如果您正确地迭代您的dsf数组并单独打印每个元素，您将获得有效的输出。如下所示：|392350|for(String+str+:+dsf){
+++System.out.println(str);
}|code-block|syntax|javascript|392351|您可能想要一次打印完整的Array对象，因为这会给出一个对象内存地址，就像您在问题中得到的那样。如果看到这种输出，则需要为要打印的对象提供一个toString()方法。|392352|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/1.5.0/docs/api/java/util/ArrayList.html#toArray%2528%2529^0|2L|2B|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|6|7|Y|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|Z|8|@]|9|@]|C|$I|J]]|$1|K|3|L|5|6|7|10|8|@]|9|@]|C|$]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

What you did with the iteration is not wrong from what I can make of it based on the question. It gives you a valid array of String objects. Like mentioned in another answer it is however easier to use the toArray() method available for the ArrayList object => <a href="http://docs.oracle.com/javase/1.5.0/docs/api/java/util/ArrayList.html#toArray%28%29" rel="nofollow">http://docs.oracle.com/javase/1.5.0/docs/api/java/util/ArrayList.html#toArray%28%29</a>

Just a side note. If you would iterate your dsf array properly and print each element on its own you would get valid output. Like this:

<pre><code>for(String str : dsf){
 System.out.println(str);
}
</code></pre>

What you probably tried to do was print the complete Array object at once since that would give an object memory address like you got in your question. If you see that kind of output you need to provide a toString() method for the object you're printing.

blocks|key|392378|text|package+com.v4common.shared.beans.audittrail;

import+java.util.ArrayList;
import+java.util.List;

public+class+test1+{
++++public+static+void+main(String+arg[]){
++++++++List<String>+list+=+new+ArrayList<String>();
++++++++list.add("abcd#xyz");
++++++++list.add("mnop#qrs");

++++++++Object[]+s+=+list.toArray();
++++++++String[]+s1=+new+String[list.size()];
++++++++String[]+s2=+new+String[list.size()];

++++++++for(int+i=0;i<s.length;i%2B%2B){
++++++++++++if(s[i]+instanceof+String){
++++++++++++++++String+temp+=+(String)s[i];
++++++++++++++++if(temp.contains("#")){
++++++++++++++++++++String[]+tempString+=+temp.split("#");
++++++++++++++++++++for(int+j=0;j<tempString.length;j%2B%2B)+{
++++++++++++++++++++++++s1[i]+=+tempString[0];
++++++++++++++++++++++++s2[i]+=+tempString[1];
++++++++++++++++++++}

++++++++++++++++}
++++++++++++}+++
++++++++}
++++++++System.out.println(s1.length);
++++++++System.out.println(s2.length);
++++++++System.out.println(s1[0]);
++++++++System.out.println(s1[1]);
++++}
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|392379|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>package com.v4common.shared.beans.audittrail;

import java.util.ArrayList;
import java.util.List;

public class test1 {
 public static void main(String arg[]){
 List&lt;String&gt; list = new ArrayList&lt;String&gt;();
 list.add("abcd#xyz");
 list.add("mnop#qrs");

 Object[] s = list.toArray();
 String[] s1= new String[list.size()];
 String[] s2= new String[list.size()];

 for(int i=0;i&lt;s.length;i++){
 if(s[i] instanceof String){
 String temp = (String)s[i];
 if(temp.contains("#")){
 String[] tempString = temp.split("#");
 for(int j=0;j&lt;tempString.length;j++) {
 s1[i] = tempString[0];
 s2[i] = tempString[1];
 }

 }
 } 
 }
 System.out.println(s1.length);
 System.out.println(s2.length);
 System.out.println(s1[0]);
 System.out.println(s1[1]);
 }
}
</code></pre>

blocks|key|608580|text|以下是针对给定场景的解决方案-|type|unstyled|depth|inlineStyleRanges|entityRanges|data|608581|List<String>ls+=+new+ArrayList<String>();
++++ls.add("dfsa#FSDfsd");
++++ls.add("dfsdaor#ooiui");
++++String[]+firstArray+=+new+String[ls.size()];++++
+firstArray+=ls.toArray(firstArray);
String[]+secondArray+=+new+String[ls.size()];
for(int+i=0;i<ls.size();i%2B%2B){
secondArray[i]=firstArray[i].split("#")[0];
firstArray[i]=firstArray[i].split("#")[1];
}+|code-block|syntax|javascript|608582|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|$]]

Here is the solution for you given scenario - 

<pre><code>List&lt;String&gt;ls = new ArrayList&lt;String&gt;();
 ls.add("dfsa#FSDfsd");
 ls.add("dfsdaor#ooiui");
 String[] firstArray = new String[ls.size()]; 
 firstArray =ls.toArray(firstArray);
String[] secondArray = new String[ls.size()];
for(int i=0;i&lt;ls.size();i++){
secondArray[i]=firstArray[i].split("#")[0];
firstArray[i]=firstArray[i].split("#")[1];
} 
</code></pre>

blocks|key|608613|text|import+java.util.*;
public+class+arrayList+{
++++public+static+void+main(String[]+args)+{
++++++++Scanner+sc=new+Scanner(System.in);
++++++++ArrayList<String+>+x=new+ArrayList<>();
++++++++//inserting+element
++++++++x.add(sc.next());
++++++++x.add(sc.next());
++++++++x.add(sc.next());
++++++++x.add(sc.next());
++++++++x.add(sc.next());
+++++++++//to+show+element
+++++++++System.out.println(x);
++++++++//converting+arraylist+to+stringarray
+++++++++String[]a=x.toArray(new+String[x.size()]);
++++++++++for(String+s:a)
+++++++++++System.out.print(s%2B"+");
++}

}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|608614|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>import java.util.*;
public class arrayList {
 public static void main(String[] args) {
 Scanner sc=new Scanner(System.in);
 ArrayList&lt;String &gt; x=new ArrayList&lt;&gt;();
 //inserting element
 x.add(sc.next());
 x.add(sc.next());
 x.add(sc.next());
 x.add(sc.next());
 x.add(sc.next());
 //to show element
 System.out.println(x);
 //converting arraylist to stringarray
 String[]a=x.toArray(new String[x.size()]);
 for(String s:a)
 System.out.print(s+" ");
 }

}
</code></pre>

blocks|key|608766|text|这就是您想要的正确答案，我已经在netbeans上运行了这个解决方案|type|unstyled|depth|inlineStyleRanges|entityRanges|data|608767|ArrayList+a=new+ArrayList();
a.add(1);
a.add(3);
a.add(4);
a.add(5);
a.add(8);
a.add(12);

int+b[]=+new+int+[6];
++++++++Integer+m[]+=+new+Integer[a.size()];//***Very+important+conversion+to+array*****
++++++++m=(Integer[])+a.toArray(m);
for(int+i=0;i<a.size();i%2B%2B)
{
++++b[i]=m[i];+
++++System.out.println(b[i]);
}+++
++++System.out.println(a.size());|code-block|syntax|javascript|608768|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|$]]

This is the right answer you want and this solution i have run my self on netbeans

<pre><code>ArrayList a=new ArrayList();
a.add(1);
a.add(3);
a.add(4);
a.add(5);
a.add(8);
a.add(12);

int b[]= new int [6];
 Integer m[] = new Integer[a.size()];//***Very important conversion to array*****
 m=(Integer[]) a.toArray(m);
for(int i=0;i&lt;a.size();i++)
{
 b[i]=m[i]; 
 System.out.println(b[i]);
} 
 System.out.println(a.size());
</code></pre>

blocks|key|392642|text|这可以使用stream来完成：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|392643|List<String>+stringList+=+Arrays.asList("abc#bcd",+"mno#pqr");
++++List<String[]>+objects+=+stringList.stream()
+++++++++++++++++++++++++++++++++++++++.map(s+->+s.split("#"))
+++++++++++++++++++++++++++++++++++++++.collect(Collectors.toList());|code-block|syntax|javascript|392644|返回值将是拆分字符串数组。这避免了将arraylist转换为数组并执行操作。|392645|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|$]]

This can be done using stream:

<pre><code>List&lt;String&gt; stringList = Arrays.asList("abc#bcd", "mno#pqr");
 List&lt;String[]&gt; objects = stringList.stream()
 .map(s -&gt; s.split("#"))
 .collect(Collectors.toList());
</code></pre>

The return value would be arrays of split string. 
This avoids converting the arraylist to an array and performing the operation.

blocks|key|392724|text|public+T[]+toArray(T[]+a)+-在此方法中，我们创建一个大小为arraylist的数组，并将其作为参数传递，它将返回arraylist的元素数组|type|unstyled|depth|inlineStyleRanges|entityRanges|data|392725|++++String[]+arr+=+new+String[list.size()];+

++++arr+=+list.toArray(arr);|code-block|syntax|javascript|392726|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|$]]

public T[] toArray(T[] a) - In this method we create a array with size of arraylist and pass it as argument and it will return array of element of arraylist
<pre><code> String[] arr = new String[list.size()]; 

 arr = list.toArray(arr);
</code></pre>

I have an ArrayList with values like "abcd#xyz" and "mnop#qrs". I want to convert it into an Array and then split it with # as delimiter and have abcd,mnop in an array and xyz,qrs in another array. I tried the following code:

<pre><code>String dsf[] = new String[al.size()]; 
for(int i =0;i&lt;al.size();i++){
 dsf[i] = al.get(i);
}
</code></pre>

But it failed saying "Ljava.lang.String;@57ba57ba"

Converting ArrayList to Array in java

我有一个值为"abcd#xyz“和"mnop#qrs”的ArrayList。我想把它转换成一个数组，然后用#作为分隔符将它拆分，并将abcd，mnop放在一个数组中，将xyz，qrs放在另一个数组中。我尝试了以下代码：String dsf[] = new String[al.size()]; for(int i =0;i<al.size();i++){ dsf[i] =

问在java中将ArrayList转换为数组
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