Java Beginner Having Trouble Using Integer.parseInt for Converting a String into an integer

Question

I am creating a batch program that reads an external input file that contains

credit card numbers and then sums the digits(from right to left), starting at the

last digit and skipping every other digit. I have to treat the input as a String and not

numeric. My code looks like:

          int oddSum = 0, digit = 0;
            for(int odd = (cardNumber.length() - 1); odd >= 0; odd = odd - 2)
            {
                digit = Integer.parseInt(cardNumber);
                oddSum += digit;
            }

But String cardNumber is the whole String, while I just need the single digit. Say:

            cardNumber = "4388576018402626";

How can I get, let's say "6", turned into an integer?

Yanick Rochon · Accepted Answer · 2011-03-05 04:05:18Z

3

You don't even need to parse anything if you only need to treat one digit at a time :

String cardNumber = "4388576018402626";
int d = cardNumber.charAt(0) - '0';  // = 4

But if you insist on using Integer.parseInt(), use the String.substring(begin, end) method :

int d = Integer.parseInt(cardNumber.substring(0, 1));  // = 4

** UPDATE **

Or perhaps this even better solution :

char[] chars = cardNumber.toCharArray();
int oddSum = 0, digit = 0, len = chars.length - 1;
for(int odd = len; odd >= 0; odd -= 2)
{
    if (Character.isDigit(chars[odd])) {
       digit = Character.getNumericValue(chars[odd]);
       oddSum += digit;
    } //else {
      // do something... throw an exception, whatever here 
    //}
}

edited Mar 5, 2011 at 4:05

answered Mar 5, 2011 at 3:38

Yanick Rochon

53.8k30 gold badges146 silver badges231 bronze badges

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

6 Comments

rlibby Over a year ago

There is some value to sticking with parseInt. If cardNumber happens to contain something besides decimal digits, parseInt will throw an exception, while the above character encoding hack will just silently produce junk.

Yanick Rochon Over a year ago

@rlibby, I'm not too fund of using try..catch blocks in simple validation if possible. For example, using charAt(i) - '0' should produce a value between 0 and 9; any other value should be considered as an invalid input character. No need to throw an exception for that. Plus, the charAt() method is faster than using substring() and parseInt() ... or worst! using pasreInt(Stirng.valueOf(char)) !!

rlibby Over a year ago

@Yanick Rochon, the exception parseInt throws in unchecked (java.lang.NumberFormatException), and, yes, there is a speed-safety trade off.

Yanick Rochon Over a year ago

@rlibby, I still don't see any benefit in Integer.parseInt() (see update)

rlibby Over a year ago

@Yanick Rochon, your update still just fails silently. If cardNumber is "juicy mangoes" then your update yields 0 for oddSum. This is fine if you have already validated the input, but obviously bogus if not. On the other hand you could just let parseInt do the validating.

|

rlibby · Accepted Answer · 2011-03-05 03:39:31Z

2

How can I get, let's say "6", turned into an integer?

Integer.parseInt("6");

But String cardNumber is the whole String, while I just need the single digit.

So then just get a single digit:

String cardNumber = "4388576018402626";
String digit = cardNumber.substring(i, i + 1);

answered Mar 5, 2011 at 3:39

rlibby

6,05122 silver badges25 bronze badges

Comments

ide · Accepted Answer · 2011-03-05 03:37:22Z

1

You should iterate through the string as if it were an array of individual characters; use cardNumber.toCharArray().

int oddSum = 0, digit = 0;
char[] cardNumberDigits = cardNumber.toCharArray();
for(int odd = cardNumberDigits.length - 1; odd >= 0; odd = odd - 2) {
    digit = Integer.parseInt(String.valueOf(cardNumberDigits[odd]));
    oddSum += digit;
}

answered Mar 5, 2011 at 3:37

ide

20.9k6 gold badges72 silver badges113 bronze badges

2 Comments

tscudder16 Over a year ago

But then how do I get the char from the array turned into an integer?

ide Over a year ago

Use String.valueOf or Character.toString to convert the char to a String and then use Integer.parseInt. The key is to make your code read cleanly.

evanskis · Accepted Answer · 2011-03-05 03:38:26Z

0

The charAt() accessor method allows you to get a single character at an index.

answered Mar 5, 2011 at 3:38

evanskis

511 silver badge6 bronze badges

Comments

pjama · Accepted Answer · 2011-03-05 03:39:45Z

0

You need to grab a substring of cardNumber

String cardNumber = "4388576018402626";
int start = cardNumber.length()-2;
int end = cardNumber.length()-1;
String subString = cardNumber.substring(start, end);
int numberSix = Integer.parseInt(subString);

answered Mar 5, 2011 at 3:39

pjama

3,0443 gold badges29 silver badges27 bronze badges

Collectives™ on Stack Overflow

Java Beginner Having Trouble Using Integer.parseInt for Converting a String into an integer

5 Answers 5

6 Comments

Comments

2 Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

5 Answers 5

6 Comments

Comments

2 Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related