Java Array Indexing

Question

    class anEvent{ 
  String number;
  String dueTime;
 }



public static void main(String args[]) {
      int x = args.length / 2;
      int y = args.length;
      anEvent [] order = new anEvent [x];
      for(int i=0; i<x; i++){
       if(i==0){
        order[i].number = args[0]; //Line(#)
        order[i].dueTime = args[1];
       } else if ( i % 2 == 0){
       order[i].number = args[i];
       order[i].dueTime = args[i];
       } else if ( i % 2 != 0){
        order[i].number = args[i+1];
        order[i].dueTime = args[i+1];
       } else if ( i == x -1){
        order[i].number = args[x-1];
        order[i].dueTime = args[x-1];
       }

      }

Java complains that a Null Pointer exceptuion is present at line # in the above snippet.

What's the matter?

ps: I know that the snippet can be cleaned up but there should be no problem at all on line #

Answer 1

When an array is created, all the array elements are null. In your case, you need to fill the array with new anEvent() instances.

Answer 2

Make the first line of your for-loop:

order[i] = new anEvent();

As is, you are not initializing anything in the array (they're all null), so when you try to access the fields you get that exception.

Answer 3

You hasn't created any anEvent instance, defining an array (order[]) you are not creating default values for it.

and also there are more simple array for your case:

List events = new ArrayList(x);
for(int i=0; i<y; i+=2){
  anEvent event = new anEvent();
  anEvent.number = args[i];
  anEvent.dueTime = args[i+1];
  events.add(event);
}
anEvent[] order = events.toArray();

Answer 4

Since you mention that it "can be cleaned up", I took the liberty of so doing:

public class Thing {
    private String number;
    private String dueTime;

    public Thing(String number, String dueTime) {
        this.number = number;
        this.dueTime = dueTime;
    }

    public static void main(String args[]) {
        int x = args.length / 2;
        Thing[] order = new Thing[x];
        for (int i = 0; i < x; i++) {
            if (i == 0) {
                order[i] = new Thing(args[0], args[1]);
            } else if (i % 2 == 0) {
                order[i] = new Thing(args[i], args[i]);
            } else if (i % 2 != 0) {
                order[i] = new Thing(args[i + 1], args[i + 1]);
            }
        }
    }
}

"anEvent" doesn't conform to the capitalized camel-case Java standard, so I renamed it. "Thing" isn't particularly meaningful, but there isn't much to work with here. The final else if clause can never be reached, because i % 2 either is or is not equal to zero, so I dropped it. And, of course, I'm creating new Things, which avoids the problem of the nulls. Enjoy.

Answer 5

NullPointerException means that you attempted to add a value or execute a method to something that turn out to be a null.

In Java object references can have assigned ... well, objects and null

When they have null assigned this exception is thrown:

Object o = null;
o.toString(); // <- NullPointerException ( think of null.toString() )

Arrays, are objects also. When you create an array with a size, all the "boxes" inside the array contain null as reference.

So:

Object[] array = new Object[10];

Creates something similar to the following:

 [null,null,null,null,null,null,null,null,null,null]

That's why, when you execute:

array[0].toString(); // or  order[i].number in your specific example... 

You get that exception, because the effect is exactly the same as:

null.toString();  // or null.number  <-- NullPointerException.

To solve this problem, you have to assign a valid object reference to that position into the array:

for(int i=0; i<x; i++){
    order[i] = new anEvent();
    ...
    ...

I hope this helps.

Final note. In Java classes names start with upper case, so your class should've really be:

class AnEvent {
....

And finally, most of the java source code is indented using 4 spaces.