Converting ArrayList to Array in java

Question

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:

String dsf[] = new String[al.size()];              
for(int i =0;i<al.size();i++){
  dsf[i] = al.get(i);
}

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

POst the definition of your ArrayList and the complete error message — Guillaume Polet, Mar 29 '12 at 16:18
What do you mean by "it failed saying ..."? That looks like the result of printing `dsf.toString()`... — Jon Skeet, Mar 29 '12 at 16:19
Do you want an array of arrays of Strings, i.e. a 2 dimensional array? — Adam, Mar 29 '12 at 16:20
possible duplicate of [Convert ArrayList containing Strings to an array of Strings in Java?](http://stackoverflow.com/questions/4042434/convert-arraylist-containing-strings-to-an-array-of-strings-in-java) — Nateowami, Dec 04 '14 at 09:45

score 123 · Answer 1 · answered Mar 29 '12 at 16:18

123

You don't need to reinvent the wheel, here's the toArray() method:

String []dsf = new String[al.size()];
al.toArray(dsf);

answered Mar 29 '12 at 16:18

talnicolas

13,437
7
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ArrayList a1 = new ArrayList(); for (Entry entry : outmap.entrySet()) { if(entry.getKey().indexOf("Data") != -1){ exports.add(entry.getValue()); } } – Shruthi Mar 29 '12 at 16:24
There's also this form 'http://docs.oracle.com/javase/1.5.0/docs/api/java/util/ArrayList.html#toArray(T[])' for where you want to provide a specificly typed array. – wmorrison365 Mar 29 '12 at 16:29

score 59 · Answer 2 · edited Dec 15 '12 at 18:37

59

List<String> list=new ArrayList<String>();
list.add("sravan");
list.add("vasu");
list.add("raki");
String names[]=list.toArray(new String[list.size()])

edited Dec 15 '12 at 18:37

Aviram Segal

10,792
3
38
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answered Dec 15 '12 at 18:18

Sravan Kumar Limbadri

591
4
2

score 8 · Answer 3 · answered Jan 22 '17 at 20:10

8

List<String> list=new ArrayList<String>();
list.add("sravan");
list.add("vasu");
list.add("raki"); 
String names[]=list.toArray(new String[0]);

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.

answered Jan 22 '17 at 20:10

nandy

81
1
3

1

I think `new String[0]` redundant because you always have the list. So you can determine the size. – ישו אוהב אותך Mar 03 '17 at 03:30
1

You can/should replace `new String[0]` with `new String[list.size()]` – ᴛʜᴇᴘᴀᴛᴇʟ May 16 '17 at 19:53

score 2 · Answer 4 · answered Jan 12 '16 at 12:08

2

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+" ");
  }

}

answered Jan 12 '16 at 12:08

Satyendra Jaiswal

17
2

1

This doesn't do the requested split. – rghome Jan 12 '16 at 12:36
provide description for an answer if you want others to understand and maybe upvote it – insomniac Jan 12 '16 at 13:55
This approach is the same as the one in [this answer posted a few years earlier](https://stackoverflow.com/a/13894859/1711796). – Bernhard Barker Nov 16 '17 at 09:37

score 2 · Answer 5 · answered Jun 08 '16 at 10:31

2

String[] values = new String[arrayList.size()];
        for (int i = 0; i < arrayList.size(); i++) {
            values[i] = arrayList.get(i).type;
        }

answered Jun 08 '16 at 10:31

Anand

79
1
8

Here arrayList.get(i).type is a one of the value in the arrayList. You can get it anyway. – Anand Jun 08 '16 at 10:33

score 1 · Answer 6 · answered Aug 26 '17 at 15:05

1

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

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++)
{
    b[i]=m[i]; 
    System.out.println(b[i]);
}   
    System.out.println(a.size());

answered Aug 26 '17 at 15:05

Abhinav khosla

11
1

1

How is your answer better when the existing one and now only worse due to the use of raw-types? – Tom Aug 26 '17 at 15:19

score 1 · Answer 7 · answered Mar 29 '12 at 16:30

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 => http://docs.oracle.com/javase/1.5.0/docs/api/java/util/ArrayList.html#toArray%28%29

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:

for(String str : dsf){
   System.out.println(str);
}

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.

score 1 · Answer 8 · answered Mar 29 '12 at 16:39

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++){
            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++) {
                        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]);
    }
}

score 1 · Answer 9 · answered Mar 29 '12 at 17:08

Here is the solution for you given scenario -

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++){
secondArray[i]=firstArray[i].split("#")[0];
firstArray[i]=firstArray[i].split("#")[1];
}

score 0 · Answer 10 · edited May 24 '18 at 11:42

This can be done using stream:

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

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

score 0 · Answer 11 · answered Jan 25 '21 at 17:23

We can convert ararylist to array using 3 mrthod

public Object[] toArray() - it will return array of object

Object[] array = list.toArray();
public T[] toArray(T[] a) - In this way we will create array and toArray Take it as argument then return it
```
   String[] arr = new String[list.size()]; 
    arr = list.toArray(arr);
```
Public get() method;

Iterate ararylist and one by one add element in array.

For more details for these method Visit Java Vogue

score 0 · Answer 12 · answered Jan 27 '21 at 08:56

0

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

    String[] arr = new String[list.size()]; 

    arr = list.toArray(arr);

answered Jan 27 '21 at 08:56

php king

105
2
2
12

score 0 · Answer 13 · answered Jun 01 '22 at 09:31

0

NameOfArray.toArray(new String[0])

This will convert ArrayList to Array in java

answered Jun 01 '22 at 09:31

akshay shetty

174
3
10

Converting ArrayList to Array in java

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