1296

How might I convert an ArrayList<String> object to a String[] array in Java?

1
  • 5
    Have made this answer with an updated approach with JDK-11 introducing a new an equally performant API to toArray(T[]) and similar in syntax to Stream.toArray.
    – Naman
    Commented Jul 26, 2018 at 18:40

17 Answers 17

2098
List<String> list = ..;
String[] array = list.toArray(new String[0]);

For example:

List<String> list = new ArrayList<String>();
//add some stuff
list.add("android");
list.add("apple");
String[] stringArray = list.toArray(new String[0]);

The toArray() method without passing any argument returns Object[]. So you have to pass an array as an argument, which will be filled with the data from the list, and returned. You can pass an empty array as well, but you can also pass an array with the desired size.

Important update: Originally the code above used new String[list.size()]. However, this blogpost reveals that due to JVM optimizations, using new String[0] is better now.

5
  • It shows this warning in logcat: Converting to string: TypedValue{t=0x12/d=0x0 a=3 r=0x7f050009}
    – Adil Malik
    Commented Jun 19, 2013 at 12:40
  • @ThorbjørnRavnAndersen: the size of the array doesn't make any difference functionally but if you pass it in with the right size, you save the JVM the work of resizing it.
    – sferencik
    Commented Oct 26, 2015 at 10:57
  • 77
    Turns out that providing a zero-length array, even creating it and throwing it away, is on average faster than allocating an array of the right size. For benchmarks and explanation see here: shipilev.net/blog/2016/arrays-wisdom-ancients Commented Jan 21, 2016 at 8:29
  • @Nyerguds Can you explain how java's fake generics disallow such function?
    – Lyuboslav
    Commented Nov 1, 2017 at 17:34
  • 3
    @lyuboslavkanev The problem is that having the generic type in Java isn't enough to actually create objects based on that type; not even an array (which is ridiculous, because that should work for any type). All that can be done with it, as far as I can see, is casting. In fact, to even create objects of the type, you need to have the actual Class object, which seems to be completely impossible to derive from the generic type. The reason for this, as I said, is that the whole construction is fake; it's all just stored as Object internally.
    – Nyerguds
    Commented Nov 3, 2017 at 7:52
274

An alternative in Java 8:

String[] strings = list.stream().toArray(String[]::new);

Java 11+:

String[] strings = list.toArray(String[]::new);
8
  • 28
    Or any benefits? Commented Apr 2, 2016 at 13:50
  • 1
    I would prefer that syntax, but IntelliJ displays a compiler error with that, complains "T[] is not a functional interface."
    – Glen Mazza
    Commented Jul 25, 2016 at 16:11
  • 3
    @GlenMazza you can only use toArray on a Stream object. This compilation error may occur if you reduce the stream using a Collectors and then try to apply toArray. Commented Nov 16, 2016 at 7:29
  • I don't get it. String[]::new is equivalent to () -> new String[]. But the given interface method expects an integer. Shouldn't it be (list.size()) -> new String[], so String[list.size()]::new ? Commented Dec 11, 2020 at 14:31
  • 4
    @JohnStrood String[]::new is equivalent to i -> new String[i]. See stackoverflow.com/questions/29447561/… Commented Feb 11, 2021 at 6:37
63

Starting from Java-11, one can use the API Collection.toArray(IntFunction<T[]> generator) to achieve the same as:

List<String> list = List.of("x","y","z");
String[] arrayBeforeJDK11 = list.toArray(new String[0]);
String[] arrayAfterJDK11 = list.toArray(String[]::new); // similar to Stream.toArray
0
49

You can use the toArray() method for List:

ArrayList<String> list = new ArrayList<String>();

list.add("apple");
list.add("banana");

String[] array = list.toArray(new String[list.size()]);

Or you can manually add the elements to an array:

ArrayList<String> list = new ArrayList<String>();

list.add("apple");
list.add("banana");

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

for (int i = 0; i < list.size(); i++) {
    array[i] = list.get(i);
}

Hope this helps!

34
ArrayList<String> arrayList = new ArrayList<String>();
Object[] objectList = arrayList.toArray();
String[] stringArray =  Arrays.copyOf(objectList,objectList.length,String[].class);

Using copyOf, ArrayList to arrays might be done also.

0
11

In Java 8:

String[] strings = list.parallelStream().toArray(String[]::new);
2
  • 13
    This is really already contained in this answer. Perhaps add it as an edit to that answer instead of as an entirely new one.
    – River
    Commented Feb 5, 2016 at 16:17
  • 16
    Why parallelStream() instead of simply stream()?
    – Yoory N.
    Commented Feb 14, 2018 at 12:44
8

In Java 8, it can be done using

String[] arrayFromList = fromlist.stream().toArray(String[]::new);
7

You can use Iterator<String> to iterate the elements of the ArrayList<String>:

ArrayList<String> list = new ArrayList<>();
String[] array = new String[list.size()];
int i = 0;
for (Iterator<String> iterator = list.iterator(); iterator.hasNext(); i++) {
    array[i] = iterator.next();
}

Now you can retrive elements from String[] using any Loop.

1
  • I downvoted because: 1. no use of generics which force you into 2. using .toString() where no explicit cast would be needed, 3. you don't even increment i, and 4. the while would be better off replaced by a for. Suggested code: ArrayList<String> stringList = ... ; String[] stringArray = new String[stringList.size()]; int i = 0; for(Iterator<String> it = stringList.iterator(); it.hasNext(); i++) { stringArray[i] = it.next(); } Commented Aug 28, 2017 at 9:05
6

If your application is already using Apache Commons lib, you can slightly modify the accepted answer to not create a new empty array each time:

List<String> list = ..;
String[] array = list.toArray(ArrayUtils.EMPTY_STRING_ARRAY);

// or if using static import
String[] array = list.toArray(EMPTY_STRING_ARRAY);

There are a few more preallocated empty arrays of different types in ArrayUtils.

Also we can trick JVM to create en empty array for us this way:

String[] array = list.toArray(ArrayUtils.toArray());

// or if using static import
String[] array = list.toArray(toArray());

But there's really no advantage this way, just a matter of taste, IMO.

6

Generics solution to covert any List<Type> to String []:

public static  <T> String[] listToArray(List<T> list) {
    String [] array = new String[list.size()];
    for (int i = 0; i < array.length; i++)
        array[i] = list.get(i).toString();
    return array;
}

Note You must override toString() method.

class Car {
  private String name;
  public Car(String name) {
    this.name = name;
  }
  public String toString() {
    return name;
  }
}
final List<Car> carList = new ArrayList<Car>();
carList.add(new Car("BMW"))
carList.add(new Car("Mercedes"))
carList.add(new Car("Skoda"))
final String[] carArray = listToArray(carList);
1
  • If generics has a meaning, wouldn't it rather be "convert any List<Type> to Type[]"? Commented Jun 6, 2021 at 10:01
6
    List<String> list = new ArrayList<>();
    list.add("a");
    list.add("b");
    list.add("c");
    String [] strArry= list.stream().toArray(size -> new String[size]);

Per comments, I have added a paragraph to explain how the conversion works. First, List is converted to a String stream. Then it uses Stream.toArray to convert the elements in the stream to an Array. In the last statement above "size -> new String[size]" is actually an IntFunction function that allocates a String array with the size of the String stream. The statement is identical to

IntFunction<String []> allocateFunc = size -> { 
return new String[size];
};   
String [] strArry= list.stream().toArray(allocateFunc);
1
  • 5
    What value does this answer add? It appears to just repeat other answers without explaining why it answers the question.
    – Richard
    Commented Feb 21, 2019 at 7:56
5
List <String> list = ...
String[] array = new String[list.size()];
int i=0;
for(String s: list){
  array[i++] = s;
}
1
  • 13
    This works, but isn't super efficient, and duplicates functionality in the accepted answer with extra code. Commented Feb 8, 2013 at 15:26
5

In Java 11, we can use the Collection.toArray(generator) method. The following code will create a new array of strings:

List<String> list = List.of("one", "two", "three");
String[] array = list.toArray(String[]::new)

from java.base's java.util.Collection.toArray().

0
5

In case some extra manipulation of the data is desired, for which the user wants a function, this approach is not perfect (as it requires passing the class of the element as second parameter), but works:

import java.util.ArrayList;
import java.lang.reflect.Array;

public class Test {
  public static void main(String[] args) {
    ArrayList<Integer> al = new ArrayList<>();
    al.add(1);
    al.add(2);
    Integer[] arr = convert(al, Integer.class);
    for (int i=0; i<arr.length; i++)
      System.out.println(arr[i]);
  }

  public static <T> T[] convert(ArrayList<T> al, Class clazz) {
    return (T[]) al.toArray((T[])Array.newInstance(clazz, al.size()));
  }
}
2

You can convert List to String array by using this method:

 Object[] stringlist=list.toArray();

The complete example:

ArrayList<String> list=new ArrayList<>();
    list.add("Abc");
    list.add("xyz");

    Object[] stringlist=list.toArray();

    for(int i = 0; i < stringlist.length ; i++)
    {
          Log.wtf("list data:",(String)stringlist[i]);
    }
1
  • output type is wrong. Commented Jun 3, 2022 at 8:43
1
private String[] prepareDeliveryArray(List<DeliveryServiceModel> deliveryServices) {
    String[] delivery = new String[deliveryServices.size()];
    for (int i = 0; i < deliveryServices.size(); i++) {
        delivery[i] = deliveryServices.get(i).getName();
    }
    return delivery;
}
0

An alternate one-liner method for primitive types, such as double, int, etc.:

List<Double> coordList = List.of(3.141, 2.71);
double[] doubleArray = coordList.mapToDouble(Double::doubleValue).toArray();
List<Integer> coordList = List.of(11, 99);
int[] intArray = coordList.mapToInt(Integer::intValue).toArray();

and so on...

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