I get null values while adding on the same index of an array

Question

Hey i was trying to recognize numbers from a string and convert them into a String array, but with this code what i get is:

IE: [null1, null2, exc..]

I dont know how to take out this null value.
I also would like to find a way to do this, without an array, because i should increase its length manually.

Any tips? Thanks!

public class ProvaToArray {
    public static void main(String[] x) {
        String s = "1-2-3-4-lorem23ip567um1";
        String[] ris = toArray(s);
        System.out.println(Arrays.toString(ris)); //should printout  [1, 2, 3, 4, 23, 567, 1]
    }

    public static String[] toArray(String s){
        String[] l = new String[10];
        int count = 0;
        for(int i = 0; i < s.length() - 1; i++){
            if(estNumber(s.charAt(i) + "")){
                l[count] += "" + s.charAt(i);
                if(!estNumber(s.charAt(i+1) + "")){
                    count++;
                }
            }
            if(i+1 == s.length()-1){
                if(estNumber(s.charAt(i+1) + "")){
                    l[count] = "" + s.charAt(i+1);
                }
            }
        }
        return l;
    }

    public static boolean estNumber(String i){
        if(i.matches("^-?\\d+$"))
            return true;
        else
            return false;
    }
}

Eran · Accepted Answer · 2015-07-23 16:47:44Z

4

When you initialize your String array :

String[] l = new String[10];

Each index contains null.

When you append a String to that null, you get the String "null" append to your String.

If you change :

l[count] += "" + s.charAt(i);

to

if (l[count] == null)
    l[count] = "" + s.charAt(i);
else
    l[count] += "" + s.charAt(i);

You'll get rid of the "null" String.

I saw the code you posted as an answer. It shouldn't be posted as an answer if it doesn't answer the question. You should have added it to the question.

Anyway, your problem is accessing the i+2'th character of the String s, which leads to IndexOutOfBoundsException in the final iteration of your loop. Instead of checking the i'th, i+1'th and i+2'th character in the same iteration, you should check one character in each iteration, and use a local variable to hold the current number.

Your code can be much simpler than it currently is :

public static String[] toArray(String s)
{
    ArrayList<String> m = new ArrayList<>();
    StringBuilder currentNumber = new StringBuilder();
    for(int i = 0; i < s.length(); i++) {
        if(Character.isDigit(s.charAt(i))) {
            currentNumber.append (s.charAt(i));
        } else {
            if (currentNumber.length() > 0) {
                m.add(currentNumber.toString());
                currentNumber.setLength(0);
            }
        }
    }
    if (currentNumber.length() > 0) {
        m.add(currentNumber.toString());
    }                    
    return m.toArray(new String[m.size()]);
}

This produces the array [1, 2, 3, 4, 23, 567, 1] for the input you used.

edited Jul 23, 2015 at 16:47

answered Jul 23, 2015 at 9:46

Eran

395k57 gold badges726 silver badges793 bronze badges

Sign up to request clarification or add additional context in comments.

9 Comments

Kargzul Over a year ago

Awesome thanks! But if i dont want to use a String[], but an arrayList? how can i get rid of this?

Eran Over a year ago

@Benny If you use an ArrayList, you must either call list.add("" + s.charAt(i)) or list.set(count,list.get(count) + s.charAt(i)); depending on whether or not you already added an element to the count index of the List. You'll still need a condition.

Kargzul Over a year ago

My real problem is to add it on the same index in the arrayList.. i tried with set, add, but still cant manage it

Eran Over a year ago

@Benny list.set(count,list.get(count) + s.charAt(i)); should work, as long as list.size()>count.

Kargzul Over a year ago

it throws IndexOutOfBoundExceptions. Dont understand why

|

Collectives™ on Stack Overflow

I get null values while adding on the same index of an array

1 Answer 1

9 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

9 Comments

Your Answer

Sign up or log in

Post as a guest

Related