Following methods can be used for converting ArrayList to Array:

Method 1: Using Object[] toArray() method
Syntax:
public Object[] toArray()
- It is specified by toArray in interface Collection and interface List
- It overrides toArray in class AbstractCollection
- It returns an array containing all of the elements in this list in the correct order.
Java
import java.io.*;
import java.util.List;
import java.util.ArrayList;
class GFG {
public static void main(String[] args)
{
List<Integer> al = new ArrayList<Integer>();
al.add( 10 );
al.add( 20 );
al.add( 30 );
al.add( 40 );
Object[] objects = al.toArray();
for (Object obj : objects)
System.out.print(obj + " " );
}
}
|
Note: toArray() method returns an array of type Object(Object[]). We need to typecast it to Integer before using as Integer objects. If we do not typecast, we get compilation error.
Consider the following example:
Java
import java.io.*;
import java.util.List;
import java.util.ArrayList;
class GFG {
public static void main(String[] args)
{
List<Integer> al = new ArrayList<Integer>();
al.add( 10 );
al.add( 20 );
al.add( 30 );
al.add( 40 );
Integer[] objects = al.toArray();
for (Integer obj : objects)
System.out.println(obj);
}
}
|
Output:
19: error: incompatible types: Object[] cannot be converted to Integer[]
Integer[] objects = al.toArray();
^
1 error
It is therefore recommended to create an array into which elements of List need to be stored and pass it as an argument in toArray() method to store elements if it is big enough. Otherwise a new array of the same type is allocated for this purpose.
Method 2: Using T[] toArray(T[] a)
// Converts a list into an array arr[] and returns same.
// If arr[] is not big enough, then a new array of same
// type is allocated for this purpose.
// T represents generic.
public T[] toArray(T[] arr)
Note that the there is an array parameter and array return value. The main purpose of passed array is to tell the type of array. The returned array is of same type as passed array.
- If the passed array has enough space, then elements are stored in this array itself.
- If the passed array doesn’t have enough space, a new array is created with same type and size of given list.
- If the passed array has more space, the array is first filled with list elements, then null values are filled.
It throws ArrayStoreException if the runtime type of a is not a supertype of the runtime type of every element in this list.
Java
import java.io.*;
import java.util.List;
import java.util.ArrayList;
class GFG {
public static void main(String[] args)
{
List<Integer> al = new ArrayList<Integer>();
al.add( 10 );
al.add( 20 );
al.add( 30 );
al.add( 40 );
Integer[] arr = new Integer[al.size()];
arr = al.toArray(arr);
for (Integer x : arr)
System.out.print(x + " " );
}
}
|
Note : If the specified array is null then it will throw NullpointerException. See this for example.
Method 3: Manual method to convert ArrayList using get() method
We can use this method if we don’t want to use java in built toArray() method. This is a manual method of copying all the ArrayList elements to the String Array[].
// Returns the element at the specified index in the list.
public E get(int index)
Java
import java.io.*;
import java.util.List;
import java.util.ArrayList;
class GFG {
public static void main(String[] args)
{
List<Integer> al = new ArrayList<Integer>();
al.add( 10 );
al.add( 20 );
al.add( 30 );
al.add( 40 );
Integer[] arr = new Integer[al.size()];
for ( int i = 0 ; i < al.size(); i++)
arr[i] = al.get(i);
for (Integer x : arr)
System.out.print(x + " " );
}
}
|
Method 4: Using streams API of collections in java 8 to convert to array of primitive int type
We can use this streams() method of list and mapToInt() to convert ArrayList<Integer> to array of primitive data type int
int[] arr = list.stream().mapToInt(i -> i).toArray();
Java
import java.io.*;
import java.util.List;
import java.util.ArrayList;
class GFG {
public static void main(String[] args)
{
List<Integer> al = new ArrayList<Integer>();
al.add( 10 );
al.add( 20 );
al.add( 30 );
al.add( 40 );
int [] arr = al.stream().mapToInt(i -> i).toArray();
for ( int x : arr)
System.out.print(x + " " );
}
}
|
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