Java - Null Pointer Exception error

Question

Im getting a NullPointerException error on my code. I really dont know why it is happenning. This's only a sample of the code, i can post the full version if needed.

private static final Scanner in=new Scanner(System.in);
public static void main(String[] args) {
   int n1=lerTamanho();
   String [] vec1=new String [n1];
   readVector(vec1); -> Line 11 
}

private static int lerTamanho() {
    System.out.print("Number: ");
    int num=in.nextInt();
    while (num<=0) {
        System.out.print("Error! Number: ");
        num=in.nextInt();
    }
    return num;
}

private static void readVector(String vec[]) {
    int cont=0;
    String s;
    do  {  
        System.out.print("Name: ");
        s=in.nextLine();
        if (validate(s,vec)) { -> Line 30
            vec[cont]=s;
            cont++;
        } else {
            System.out.println("That name already exists!");
        }
    } while (cont<vec.length);
    }
private static boolean validate(String s, String vec[]) {
    boolean success=true;
    for (int i=0; i<vec.length; i++) {
        if (vec[i].equalsIgnoreCase(s)) { >-> Line 44
            return !success;
        }
    }
    return success;
}

Error I'm getting:

Exception in thread "main" java.lang.NullPointerException
at javaapplication63.JavaApplication63.validate(JavaApplication63.java:41)
at javaapplication63.JavaApplication63.lerVetor(JavaApplication63.java:30)
at javaapplication63.JavaApplication63.main(JavaApplication63.java:11)

something must have called validate passing it null for either s or vec — Neil Locketz
– Neil Locketz, Commented Nov 20, 2013 at 17:11
It would be very useful if you pointed out which line is line 41. — biziclop
– biziclop, Commented Nov 20, 2013 at 17:11
Looks like either vec or s (in the validate method) hasn't been initialized. You might want to post the code for the lerVector() method. — Josh
– Josh, Commented Nov 20, 2013 at 17:11

Konstantin Yovkov · Accepted Answer · 2013-11-20 17:11:45Z

3

By default, all the elements in the String[] are initialized with null. The NullPointerException is thrown on this line:

if (vec[i].equalsIgnoreCase(s))

In order to fix it, assign the elements with values, before validating

answered Nov 20, 2013 at 17:11

Konstantin Yovkov

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SimplyPanda · Accepted Answer · 2013-11-20 17:12:14Z

1

When you make a new array, ie String [] vec1=new String [n1];, the elements inside the array are initialized to null. Thus, when you try to access vec[i].equalsIgnoreCase(s) in

if (vec[i].equalsIgnoreCase(s)) {

you get a NPE.

answered Nov 20, 2013 at 17:12

SimplyPanda

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3 Comments

DaoWen Over a year ago

I'd change the wording here to say the elements of the array are initialized to null rather than uninitialized. Saying it's uninitialized sounds like it might be filled with garbage (like what would happen in C/C++ with freshly-allocated memory). Java always sets new memory to the "zero" value, which is null for Objects, 0 for numbers and false for Booleans.

Ricardo Almeida Over a year ago

@SimplyPanda Thank you. I missed that detail ^^

SimplyPanda Over a year ago

@Exsound We always appreciate when our answers are selected; kocko's is more complete and was faster, so I do suggest it! (Also his was accurate upon conception)

Amir Afghani · Accepted Answer · 2013-11-20 17:15:02Z

1

   String [] vec1=new String [n1];

This statement initializes the array - it does not initialize the individual buckets of the array. For that you can walk the array in a look and set them explicitly.

 for (String s : vec) {
     s = "Some String"; //or new String()
 }

answered Nov 20, 2013 at 17:15

Amir Afghani

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Comments

Floegipoky · Accepted Answer · 2013-11-20 19:46:57Z

1

Switch vec[i].equalsIgnoreCase(s) to s.equalsIgnoreCase(vec[i])

edited Nov 20, 2013 at 19:46

answered Nov 20, 2013 at 17:25

Floegipoky

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2 Comments

Ricardo Almeida Over a year ago

Can you tell me the difference between them?

Floegipoky Over a year ago

if vec[i] is null, you are trying to call a method of an object which doesn't exist. By swapping them, the method is called on s, which should always be non-null. When it gets to the portion of the array that hasn't been assigned to yet, it will essentially be s.equalsIgnoreCase(null), which will just return false. That way you don't have to worry about breaking from the loop once you reach the null values. If you wanted to keep the original order, you would need to add if(vec[i] == null){break;} right before

lkamal · Accepted Answer · 2013-11-20 17:19:16Z

0

You have created an array with below code.

String [] vec1=new String [n1];

It only creates an array with n1 elements, but all values in those elements are null.

You can try below and see.

int n1=lerTamanho();
String [] vec1=new String [n1];
for (int i=0; i < n1; i++) {
    vec1[i]= "SOME-STRING-TO-THIS-ELEMENT";
}
readVector(vec1);

NPE comes from vec[i].equalsIgnoreCase(s) where vec[i] is null.

answered Nov 20, 2013 at 17:19

lkamal

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