Java: nullpointer exception

Question

In my code I'm trying to accept the arguments and put it in the the array of objects called cl. It gives me NullPointerException.

//In my main class
public static void main(String args[]) {
    int n = Integer.parseInt(args[0]);
    cl[] Array1;  
    Array1 = new cl[n];
    cl[0].a1 = Integer.parseInt(args[1]);
    cl[0].a2 = Integer.parseInt(args[2]);
}

and

//another class declaration
public class cl {
    int a1;
    int a2;
}

I'm getting a NullPointerException, I'm able to print the command line arguments. I'm not sure what the problem is.

what is arr[] Array1;? is your code not give compile time error? — bNd
– bNd, Commented Apr 17, 2014 at 3:09
cl[0].a1 should be Array1[0].a1. Similarly for the other field access. — Sotirios Delimanolis
– Sotirios Delimanolis, Commented Apr 17, 2014 at 3:15
Please, please start using proper naming conventions. Classes start uppercase, variables start lowercase. — Brian Roach
– Brian Roach, Commented Apr 17, 2014 at 3:41
possible duplicate of java - Why is this for-loop giving a nullpointer exception? among numerous others. — Brian Roach
– Brian Roach, Commented Apr 17, 2014 at 3:43

Akash Thakare · Accepted Answer · 2014-04-17 03:19:03Z

2

 cl[]  Array1 = new cl[n];

but your object is null because you haven't initialized your object of cl[]

for example.

Array1[0]= new cl();

and than access the variables.

Array1[0].a1 and Array1[0].a2

As you are doing this

cl[0].a1 = Integer.parseInt(args[1]);

cl  is not your array variable your variable is `Array1`

edited Apr 17, 2014 at 3:19

answered Apr 17, 2014 at 3:11

Akash Thakare

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Sotirios Delimanolis · Accepted Answer · 2014-04-17 03:09:52Z

2

Initializing an array does not initialize its elements.

This

 Array1 = new cl[n];

creates an array of type cl with n elements, but those are all initialized to null since cl is a reference type.

You need to first initialize the element before accessing its field

Array1[0] = new cl();
Array1[0].a1 = ...

Try to follow Java naming conventions. Always start variable names with a lower case character. Always start class names with an upper case character.

answered Apr 17, 2014 at 3:09

Sotirios Delimanolis

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3 Comments

Jason Over a year ago

I think you mean Array1[0] = new c1(); and Array1[0].a1 = ...

Sotirios Delimanolis Over a year ago

@Jason Yeah, copy paste error from OP's snippet. Thanks.

Jason Over a year ago

I had the same problem

Jason · Accepted Answer · 2014-04-17 03:20:51Z

1

In the following lines:

c1[] Array1;  
Array1 = new cl[n];

you are creating an array in which to store references to c1.

Then you start dereferencing c1[0] when you really mean Array1[0], but you never created an instance of c1 and put it in Array1[0].

What you need to do is:

public static void main(String args[]) 
{
    int n = Integer.parseInt(args[0]);

    cl[] Array1;  

    Array1 = new cl[n];

    // don't forget to create an instance of c1 and put in array at index 0!
    Array1[0] = new c1();

    Array1[0].a1 = Integer.parseInt(args[1]);
    Array1[0].a2 = Integer.parseInt(args[2]);
}

edited Apr 17, 2014 at 3:20

answered Apr 17, 2014 at 3:11

Jason

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1 Comment

Akash Thakare Over a year ago

Jason I think it must be Array1[0] = new c1();

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