NullPointerException error while implementing Hash Table in java

Question

I am implementing Hash Table through java. When I implement the following code, I get a NullPointer Exception error but if I replace the else statement in void put(int key, int value) in class HashTable with the statement kv[key]=new KeyValue(key,value); It works! Can someone explain why is this happening? Please help!

public class KeyValue {
    int key;
    int value;

    KeyValue(int k, int v)
    {
        key = k; 
        value = v;
    }
    public int getKey(){
        return key;
    }

    public int getValue()
    {
        return value;
    }

    public void put(int k, int v)
    {
        key = k;
        value = v;
    }

}

public class HashTable{

    KeyValue[] kv;

        HashTable(){
            kv = new KeyValue[4];
            for (int i=0; i<4 ; i++)
            {    kv[i]=null;

            }
        }

    void put(int key, int value)
        {
            if((kv[key]!=null) || (key<0 && key>=4))
            {
                ;
            }
            else
            {
                kv[key].put(key,value);
            }
        }

        int get(int key)
        {
            int value;

            value=kv[key].getValue();


            return value;
        }

    public static void main(String[] a){

            HashTable h = new HashTable();
            h.put(1,2);

            System.out.println(h.get(1));

        }

}

My comment became too long, so I posted it as an answer.

SebastianH
– SebastianH

2014-04-15 10:24:29 +00:00
Commented Apr 15, 2014 at 10:24 — SebastianH
– SebastianH, Commented Apr 15, 2014 at 10:24
This is not a hash table. It is basically an array-wrapper.

user207421
– user207421

2014-04-15 10:27:49 +00:00
Commented Apr 15, 2014 at 10:27 — user207421
– user207421, Commented Apr 15, 2014 at 10:27

kai · Accepted Answer · 2014-04-15 10:23:55Z

6

You are initializing kv[key] with null. So if you do kv[key].callSomeMethod you got an NullPointerException because there is no object to call a method of.

You can initialize your KeyValues like Peter Radar said or do it in the constructor of your HashTable using the default constructor of KeyValue.

    HashTable(){
        kv = new KeyValue[4];
        for (int i=0; i<4 ; i++)
        {    
            kv[i]=new KeyValue();
        }
    }

Now an Object of type KeyValue is stored in kv[i] and you can use its methods.

edited Apr 15, 2014 at 10:23

answered Apr 15, 2014 at 10:17

kai

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1 Comment

Pablo Lozano Over a year ago

In other words: new KeyValue[4] initializes the array, but it just creates it filled with nulls, any single object must be initialized

SebastianH · Accepted Answer · 2014-04-15 10:24:02Z

1

To answer your question I believe kai and Peter Rader found the issue. But there are other things in your code that I recommend to change.

The KeyValue.put method should be renamed to set or completely dropped. The basic question is here if KeyValue should be a mutable data object or not.

In HashTable.put the your code if((kv[key]!=null) || (key<0 && key>=4)) should read if ((kv[key]!=null) || key<0 || key>=4) (besides you should of course replace 4 with a variable so you can have larger HashTables too).

Since you are trying to learn something from this I'd also recommend that you try to replace KeyValue[] kv; with a Collection.

answered Apr 15, 2014 at 10:24

SebastianH

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3 Comments

Meha Over a year ago

Thanks! I was trying to implement something very basic so that's why I used 4. Thanks for the advice though. Also ((kv[key]!=null) || key<0 || key>=4) doesn't solve my purpose as I am checking that whether the user enters a key within this range (key<0 && key>=4) or not. Because all the keys are defined in this range so it doesn't work for a key entered by the user that doesn't exist. BTW who is Peter Radar? I can't see his comment/answer, I can only see kai's answer!!!

SebastianH Over a year ago

@Meha: But key<0 && key>=4 can never be true! key can not at the same time be smaller than 0 and larger than 4 :) Its key<0 || key>=4 you are looking for. The answer by Peter seems to be retired.

Meha Over a year ago

Its this (key>0 && key<=4) what I was wanting to do :D ... there was a typo. And what you said can also be done.. its the same thing. Thanks!

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