Adding up the numbers in a String java

Question

I am trying to add up the characters in a String. But that doesn't seem to work. I am using this code:

public String createDeciOfOct (String number) {
    int result = 0;
    String[] numberArray = number.split("");
    for (String numb : numberArray) {
        if(numb != null || numb != "") {
            result += Integer.parseInt(numb);
        }
    }
    return Integer.toString(result);
}

How I am executing it is like this:

frmOctToDeci.setOnClickListener(new View.OnClickListener() {            
  @Override
  public void onClick(View v) {
    EditText edtTxt = (EditText) findViewById(R.id.octalValue);
    String value = edtTxt.getText().toString();
    String result = createDeciOfOct(value);
    TextView txtView = (TextView) findViewById(R.id.fromOctRes);
    txtView.setText(result);
  }
});

It gives me an error as:

java.lang.NumberFormatException: Invalid int: ""

Which means that there is some sort of empty string. How can I prevent this?

The logcat errors are as:

enter image description here

What you expect to do when you split with "" and you compare strings with equals , == is just for references — nachokk
– nachokk, Commented Feb 26, 2014 at 15:38
@nachokk, I want to skip it when there is no integer value there — Afzaal Ahmad Zeeshan
– Afzaal Ahmad Zeeshan, Commented Feb 26, 2014 at 15:39
and a split with no split character... what does it lead to? — Phantômaxx
– Phantômaxx, Commented Feb 26, 2014 at 15:40

Óscar López · Accepted Answer · 2014-02-26 15:50:16Z

5

Always use equals() for testing (in)equality between Strings:

!numb.equals("")

Also it might be easier to process the input as a char[], but that's just an idea:

char[] numberArray = number.toCharArray()

Then it's easy to convert a char to a digit, simply use Character.getNumericValue() on each char you wish to convert.

edited Feb 26, 2014 at 15:50

answered Feb 26, 2014 at 15:39

Óscar López

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1 Comment

Afzaal Ahmad Zeeshan Over a year ago

A nice a great solution, in both aspects! :) char[] would also be a great idea! :) However, I am a bit new to Java API that's why I am facing these troubles...

pdem · Accepted Answer · 2014-02-27 08:23:41Z

1

That is wrong, so i edited it.

        if(numb != null || numb != "")

it should be (i thought, so it's still wrong)

        if(numb != null && numb != "")

Edit: the comment is right, i just wanted to notice the && instead of || . nevermind, forget it !

so it should really be:

        if(numb != null && ! numb .equals ("") )

of course ! "".equals(numb) does the same

edited Feb 27, 2014 at 8:23

answered Feb 26, 2014 at 15:44

pdem

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2 Comments

Melquiades Over a year ago

No it should not :) When comparing strings, use equals(), not ==

pdem Over a year ago

how i could not see that basic thing i will edit, but let mi first mistake for the future!

Pierre Rust · Accepted Answer · 2014-02-26 15:50:11Z

0

First, as another answer state it, you should not use != to compare strings, but equals.

Then, split takes a regexp and is a weird way of converting a String into an array of one-character string. I think it will be better to simply take the corresponding char array and use Character.getNumericValue

char[] charArray = number.toCharArray();
for (char c: charArray ) {
    result += Character.getNumericValue(numb);
}

Be carrefull, the code above does not perform any check to make sure your initial string contains only digit characters. If there are such characters the code will not fail but will produce unexpected result.

answered Feb 26, 2014 at 15:50

Pierre Rust

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Comments

Melquiades · Accepted Answer · 2014-02-26 15:57:11Z

0

This algorithm will work:

public String createDeciOfOct (String number) {
    int result = 0;

    for (int i = 0 ; i<number.length(); i++) {
        result = result + Integer.parseInt(String.valueOf(number.charAt(i)));
    }
    return Integer.toString(result);
}

edited Feb 26, 2014 at 15:57

Melquiades

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answered Feb 26, 2014 at 15:49

Bhavit S. Sengar

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