How To Create And Initialize List Or ArrayList In One Line In Java ...
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Java Tip to create and initialize List in one line
In our post 3 ways to convert Array to ArrayList in Java, we have discussed Arrays.asList() method. You can use Arrays.asList() method to create and initialize List at the same line. java.util.Arrays class act as a bridge between Array and List in Java and by using this method you can quickly create a List from Array, which looks like creating and initializing List in one line, as shown in the below program. package test; import java.util.ArrayList; import java.util.Arrays; import java.util.List; /** * * Java program to show * How to create and initialize a List in one line in Java * Arrays.asList() our go-to method to initialize List * in the same line. * * @author http://java67.blogspot.com */ public class ArrayListDemo{ public static void main(String args[]) { //You can create and initialize Array in just one line in Java String[] coolStringArray = new String[]{"Java" , "Scala" , "Groovy"}; System.out.println(" Array : " + Arrays.toString(coolStringArray)); //Now If you want to create // an ArrayList with three elements List<String> notSoCoolStringList = new ArrayList<String>(); notSoCoolStringList.add("Java"); notSoCoolStringList.add("Scala"); notSoCoolStringList.add("Groovy"); //It took four-line to create and initialize List System.err.println(" List : " + notSoCoolStringList); //Now here is simple Java tips to // create and initialize List in one line List<String> coolStringList = Arrays.asList("Java","Scala", "Groovy"); System.err.println(" List created and initialized at same line : " + coolStringList); } } Output: Array : [Java, Scala, Groovy] List : [Java, Scala, Groovy] List created and initialized at same line : [Java, Scala, Groovy] So this was our Java tip to create and initialize a List in the same line. Just remember that Arrays.asList() return java.util.List and not ArrayList or LinkedList. Another worth noting point is that List returned by Arrays.asList() is a fixed-length list that doesn't allow you to add or remove elements from it. add() and remove() method will throw UnSupportedOperationException if you try to add or remove elements from the List. Some programs mistook this List as a read-only List, which is not because it allows set() operation which can be used to change elements in the List. Only legitimate way to createread-only Collection is Collections.unmodifiableXXX() method.You can also see these Java Collections and Stream Courses to learn more about ArrayList and other collection classes in Java.
That's all on this Java tip about how to create and initialize a List in the same line in Java. This is a great tip that can save a lot of coding time if you frequently create and initialize a List with test data. You can use this trick to quickly create ArrayList with test data for your unit testing and demos. It really saves a lot of keystrokes in long run. Other Java Collection tutorials from java67 blog Difference between Vector and ArrayList in Java ConcurrentHashMap vs HashMap in Java HashMap vs Hashtable in Java Difference between TreeSet and TreeMap in Java How to sort ArrayList in Java 
Labels: ArrayList, Java collection tutorial 3 comments:
HarinathJune 25, 2015 at 2:42 AMThere is one correction required in your article. Calling remove() method on List returned from Arrays.asList() doesn't throw any UnSupportedOperationException but calling add() method will throw the exception.
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AnonymousAugust 5, 2015 at 12:07 AMThat's not true, even calling remove() method on List returned by Arrays.asList will throw UnsupportedException as seen below :Exception in thread "main" java.lang.UnsupportedOperationException at java.util.AbstractList.remove(AbstractList.java:161) at java.util.AbstractList$Itr.remove(AbstractList.java:374) at java.util.AbstractCollection.remove(AbstractCollection.java:291)Only method which works without exception and modifies the list is set(index, value), which you can use to update the List.
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- AnitaApril 30, 2020 at 11:14 AM
Cool!
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Harinath
AnonymousFeel free to comment, ask questions if you have any doubt.
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