null pointer exception

Question

the value memanufacturer is retrieved from xml document using jdom and when this value is assigned to meman array it throws NullPointerException.

Element memanufacturer = (Element) row27.get(j9);
        meman[0] = memanufacturer.getValue();

what could be the posssible mistake.

Thanks

meman is null, StackTrace is always nice

bestsss
– bestsss

2011-03-04 09:48:50 +00:00
Commented Mar 4, 2011 at 9:48 — bestsss
– bestsss, Commented Mar 4, 2011 at 9:48

Jon Skeet · Accepted Answer · 2011-03-04 09:58:58Z

1

Assuming the exception by the second line of code, there are two obvious possibilities:

memanufacturer may be null
meman may be null

We can't tell which of these is the case, but you should be able to.

EDIT: Okay, so now we know that meman is null, that's the problem. I would suggest you use a List<String> instead:

List<String> meman = new ArrayList<String>();

...
Element memanufacturer = (Element) row27.get(j9);
meman.add(memanufacturer.getValue());

Using a List<String> instead of an array means you don't need to know the size before you start.

However, the fact that you didn't understand the error suggests you should really read a good introductory Java book before going any further with a real project. You should definitely understand how arrays, collections etc work before dealing with XML and the like. It will save you a lot of time in the long run.

edited Mar 4, 2011 at 9:58

answered Mar 4, 2011 at 9:45

Jon Skeet

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7 Comments

bestsss Over a year ago

row27.get method can throw NPE just as fine (somewhere inside), but my bet is meman for the null

technocrat Over a year ago

yes meman is initialized as null i.e String meman[] = null; and there was no issue if meman is simple string variable. but i need it to be an array as in future this piece of code is looped and i want to store all the values while looping.

Jon Skeet Over a year ago

@techocrat: Well you're trying to dereference a null variable, so of course it's blowing up. Either you need to create the array beforehand, or preferably change to using a List<String> so you don't need to know the size beforehand.

technocrat Over a year ago

its not throwing exception but its printing some peculiar values like [Ljava.lang.String;@1bf3d87 i've initialized meman as String meman[] = new String[1];

Jon Skeet Over a year ago

@technocrat: Right, so now presumably you're printing meman instead of meman[0]. As I said in my edit, learn basic Java before continuing. You're doing yourself no favours by trying to continue with your current project before you understand the fundamentals of the language.

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