Getting Exception in thread "main" java.lang.NullPointerException

Question

While running this code

public class Main
{
public int a;
public int b;
public static void main(String []args)
{
    Main []ary=new Main[26];
    int i;
    for(i=0;i<26;i++)
    {
        ary[i].a=0;
        ary[i].b=i;
    }
}
}

I am getting the following error..

Exception in thread "main" java.lang.NullPointerException
at Main.main(Main.java:11)

I created an array of objects for the same class and trying to use its instance variables

Though I searched for it, i am not able to find the mistake..

Suresh Atta · Accepted Answer · 2013-12-12 08:35:05Z

4

 Main []ary=new Main[26];

You declared array not assigned values in it.

So in memory, you array looks like Main []ary={null,null ...., null};

NullPointerException

Thrown when an application attempts to use null in a case where an object is required. These include:

Calling the instance method of a null object.
Accessing or modifying the field of a null object.
Taking the length of null as if it were an array.
Accessing or modifying the slots of null as if it were an array.
Throwing null as if it were a Throwable value.

It's like null.a which causes NullPointerException.

 for(i=0;i<26;i++)
    { 
        Main m = new Main();
        m.a =0;
        m.b =i;
        ary[i]= m;

    }

edited Dec 12, 2013 at 8:35

answered Dec 12, 2013 at 8:15

Suresh Atta

122k38 gold badges206 silver badges315 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Sachin · Accepted Answer · 2013-12-12 08:16:24Z

2

Main []ary=new Main[26];
    int i;
    for(i=0;i<26;i++)
    {
        ary[i]=new Main();
        ary[i].a=0;
        ary[i].b=i;
    }

This will work :)

answered Dec 12, 2013 at 8:16

Sachin

3,5944 gold badges25 silver badges45 bronze badges

Comments

giorashc · Accepted Answer · 2013-12-12 08:16:26Z

1

You need to create an instance for each of the array's entries in order to access it :

for(i=0;i<26;i++)
{
    ary[i] = new Main(); // Otherwise ary[i] is null and will cause an exception on the following line
    ary[i].a=0;
    ary[i].b=i;
}

answered Dec 12, 2013 at 8:16

giorashc

13.7k3 gold badges38 silver badges76 bronze badges

Comments

user2826254 · Accepted Answer · 2013-12-12 08:21:34Z

1

ary[i] is null

public class Main
{
    public int a;

    public int b;

    public static void main( String[] args )
    {
        Main[] ary = new Main[26];
        int i;
        for ( i = 0; i < 26; i++ )
        {
            ary[i]=new Main();//<---(here ary[i] was null)
            ary[i].a = 0;
            ary[i].b = i;
        }
    }
}

answered Dec 12, 2013 at 8:21

user2826254

1163 bronze badges

Comments

piet.t · Accepted Answer · 2013-12-12 08:17:31Z

0

You just created an array that can hold instances of Main, but you didn't initialize the contents, so all elements of the array are null. Do ary[i]= new Main() before assigning the values.

answered Dec 12, 2013 at 8:17

piet.t

11.9k21 gold badges45 silver badges56 bronze badges

Collectives™ on Stack Overflow

Getting Exception in thread "main" java.lang.NullPointerException

5 Answers 5

Comments

Comments

Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

5 Answers 5

Comments

Comments

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related