Null exception reading data into an array

Question

I am getting Java Exception in thread "main" java.lang.NullPointerException at test.main(test.java:) in the printf line below.

I am entering values on stdin, for example 1 2 Bus then carriage return then another line and I am ending using Ctrl-Z on Windows.

I have put print statements in and the reading in of data appears to work ok. Why am I getting null exception here?

Here is the code:

import java.util.Scanner;

public class test {
    public static class Data
    {
        public Data() {}
        public Data(int v1, int v2, String s1) { n1 = v1; n2 = v2; s = s1; } 
        public int n1; 
        public int n2;
        public String s; 
    };

    public static void main(String[] args) {
        int i = 0;
        int v1, v2;
        v1 = v2 = 0;
        String s;

        Data items[] = new Data[100];

        Scanner input = new Scanner(System.in);

        while(input.hasNext()) {
            switch(i % 3) {
                case 0: v1 = input.nextInt(); break;
            case 1: v2 = input.nextInt(); break;
                case 2: 
                    s = input.next();
                    items[i] = new Data(v1, v2, s);
                    break;
            }
            ++i;
        }
        input.close();

        for(int j = 0; j < i; ++j) {
            //on next line get Exception in thread "main" java.lang.NullPointerException at test.main(test.java:)
            System.out.printf(  "%s\t%d\t%d\n", items[j].s, items[j].n1, items[j].n2);
        }         
    }
}

Create a control point at the line in question, fire up the debugger and then look at the content of your variables, one of them is most likely null. — SirRichie
– SirRichie, Commented Jan 11, 2014 at 16:31

JB Nizet · Accepted Answer · 2014-01-11 16:32:41Z

1

Look at your code, once properly indented:

switch(i % 3) {
    case 0: 
        v1 = input.nextInt(); 
        break;
    case 1: 
        v2 = input.nextInt();
        break;
    case 2: 
        s = input.next();
        items[i] = new Data(v1, v2, s);
        break;
 }
 ++i;

This means that you only initialize items[i] if i % 3 == 2. So 2 out of three elements of the array are null.

But in the last loop, you're trying to print every element, without ever testing for null.

Side note: naming fields s, n1 and n2 is the best way to have unreadable, unmaintainable code. Give meaninful names to your classes and variables.

answered Jan 11, 2014 at 16:32

JB Nizet

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2 Comments

DaveJohnston Over a year ago

I would also like to add that you should probably use a List rather than pre-allocating an array with 100 elements. What if you have more than 100 lines of data, you will end up with an ArrayIndexOutOfBounds exception.

Angus Comber Over a year ago

Ah, how dumb, I thought it might be some esoteric Java thing.

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