Array List Null Pointer Exception - Android

Question

I am trying to create an application which retrieves the users favourite book quote however I am stuck on how to display this information back to the user. I created an ArrayList which will store all the information. However when displaying this information, I keep getting the error:

java.lang.NullPointerException

when it tries to execute the code

temp[i] = new HashMap<String,String>();

This class is shown below:

public class FavouriteQuotesActivity extends ListActivity {

    static final ArrayList<HashMap<String,String>> list = new ArrayList<HashMap<String,String>>();

    private void getFavorites() {
        DataBaseHelper myDbHelper = new DataBaseHelper(this);

        String favorites [] = myDbHelper.getFavourites();

        if(list.size() > 0)
        {
            list.removeAll(list);
        }

        for(int i = 0;i < favorites.length; i++)
        {
            String quotes = favorites[i];
            String[] quoteArray = quotes.split(":");
            HashMap<String,String> temp[] = null; 

            temp[i] = new HashMap<String,String>();
            temp[i].put("Author",(quoteArray[2]));
            temp[i].put("Quote",(quoteArray[4]));
            list.add(temp[i]);

        }


    }

Any help would be greatly appreciated.

Jon Skeet · Accepted Answer · 2012-04-17 17:30:29Z

6

Look at this code:

HashMap<String,String> temp[] = null; 
temp[i] = new HashMap<String,String>();

You've just assigned a null value to the temp variable (which would more typically have a declaration of HashMap<String, String>[] temp) - so of course you'll get a NullPointerException when you then try to access temp[i] on the very next statement.

It's not clear why you're using an array at all in that code - why aren't you just using:

HashMap<String, String> map = new HashMap<String, String>();
map.put("Author", quoteArray[2]);
map.put("Quote", quoteArray[4]);
list.add(map);

Additionally it's unclear why you're using a map at all here, given that it will always have the same two keys. Why not create a Quote class with name and author properties? Also, using a static variable here seems like a bad idea - why doesn't getFavorites create a new list on each invocation, and return it at the end of the method?

answered Apr 17, 2012 at 17:30

Jon Skeet

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3 Comments

Tudor Over a year ago

The problem is that temp should not even be an array in the first place.

Jon Skeet Over a year ago

@Tudor: Hence the second part of my answer... it's not clear that a map is really needed either.

Jon Skeet Over a year ago

@user983965: But do you understand why the change was needed? Have you considered the questions in the final paragraph?

Tudor · Accepted Answer · 2012-04-17 17:31:01Z

3

You are declaring and using temp wrong. You don't need an array of HashMap, just a single HashMap, since list is an ArrayList<HashMap>:

HashMap<String,String> temp = new HashMap<String,String>();
temp.put("Author",(quoteArray[2]));
temp.put("Quote",(quoteArray[4]));
list.add(temp);

answered Apr 17, 2012 at 17:31

Tudor

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Comments

Martijn Courteaux · Accepted Answer · 2012-04-17 17:35:46Z

3

temp doesn't point to an array. You are setting it to null.

You declared the variable temp as an array (by writing HashMap temp[]).

Without being it an actual array, you can't set any elements.

edited Apr 17, 2012 at 17:35

answered Apr 17, 2012 at 17:30

Martijn Courteaux

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7 Comments

Jon Skeet Over a year ago

It is initialized - with the null value. That's a perfectly valid initial value... it's just you can't then dereference it...

Jon Skeet Over a year ago

I'd still bicker with the details in the second part: temp is a variable. Its value is either null, or a reference to an array. When the value is null, you can't set any elements.

Martijn Courteaux Over a year ago

@Jon: Indeed. Is there a nice name for what I meant to wrote. Apparently not "initialised". A word that says it is a null pointer?

Jon Skeet Over a year ago

How about "it's being initialized with a value of null"? After all, that's what you really mean :)

Jon Skeet Over a year ago

It has a value of null, it's as simple as that. The variable is definitely assigned (so code is able to read the value) - it's just that the value is null. It was previously assigned a non-null reference, and the second assignment changes the value to a null reference.

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