Array Null Pointer Exception

Question

I'm getting a null pointer exception at line

if(names[j].compareTo(names[j+1]) > 0)

and I can't figure out why. It might have something to do with the initialization but I really don't know what it could be

public static void item3(Profitable[] movies, Scanner input) {
    int j;
    boolean flag = true;
    String temp;
    String search;
    int low = 0;
    int high;
    int mid;

    String[] names = new String[6];

    for(int i = 0; i < 5; i++) {
        names[i] = ((Movie)movies[i]).getTitle();
    }

    high = names.length - 1;
    mid = (low + high) / 2;

    while(flag)
    {
        flag = false;
        for(j = 0; j < names.length - 1; j++)
        {

            if(names[j].compareTo(names[j+1]) > 0) {
                temp = names[j];
                names[j] = names[j+1];
                names[j+1] = temp;
                flag = true;
            }
        }
    }

    System.out.print("Enter your search term: ");
    search = input.nextLine();

}

score 4 · Accepted Answer · answered Mar 30 '13 at 19:21

4

This loop:

for(int i = 0; i < 5; i++) {
    names[i] = ((Movie)movies[i]).getTitle();
}

only initializes the first five elements, but the last one does not get initialized.

answered Mar 30 '13 at 19:21

Smi

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score 0 · Answer 2 · answered Mar 30 '13 at 19:21

0

for(int i = 0; i < 5; i++) { // only 5 iterations while names contain 6 elements
    names[i] = ((Movie)movies[i]).getTitle();
}

It's always better to use i < names.length rather than an explicit integer.

for(int i = 0; i < names.length; i++)

answered Mar 30 '13 at 19:21

Eng.Fouad

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score 0 · Answer 3 · edited Mar 31 '13 at 12:27

0

You could also use i <= 5 but it's always better to use the length field.

edited Mar 31 '13 at 12:27

HamZa

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answered Mar 31 '13 at 12:02

tarunajain

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Array Null Pointer Exception

3 Answers3