What is a NumberFormatException and how can I fix it?

Question

Error Message:
Exception in thread "main" java.lang.NumberFormatException: For input string: "Ace of Clubs"
    at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)
    at java.lang.Integer.parseInt(Integer.java:580)
    at java.lang.Integer.parseInt(Integer.java:615)
    at set07102.Cards.main(Cards.java:68)
C:\Users\qasim\AppData\Local\NetBeans\Cache\8.1\executor-snippets\run.xml:53: Java returned: 1
BUILD FAILED (total time: 0 seconds)

My While Loop:

while (response != 'q' && index < 52) {
    System.out.println(cards[index]);
    int first_value = Integer.parseInt(cards[index]);
    int value = 0;
    //Add a Scanner
    Scanner scanner = new Scanner(System.in);
    System.out.println("Will the next card be higher or lower?, press q if you want to quit");
    String guess = scanner.nextLine();
    if(cards[index].startsWith("Ace")) { value = 1; }
    if(cards[index].startsWith("2")) { value = 2; }
    if(cards[index].startsWith("3")) { value = 3; }
    //checking 4-10
    if(cards[index].startsWith("Queen")){ value = 11; }
    if(cards[index].startsWith("King")){ value = 12; }
    if(guess.startsWith("h")){
        if(value > first_value){ System.out.println("You answer was right, weldone!"); } 
        else { System.out.println("You answer was wrong, try again!"); }
    } else if(guess.startsWith("l")){
        if(value < first_value) { System.out.println("You answer as right, try again!"); }
        else { System.out.println("You answer was wrong, try again!"); }
    } else { System.out.println("Your was not valid, try again!"); }
    scanner.close();            
    index++;
}//end of while loop

Lest

int first_value = Integer.parseInt(cards[index]); - you're trying to parse the string as an int, but the string is "Ace of Clubs". – Procopius

You are missing a card ... King is 13, Queen is 12 and Jack is 11, just saying ;) You should use if else because you can't have a card starting with king and also 3. Why are you using an index limit of 52 ? You are not using colors. And last thing, if you try to q, you will receive a not valid answer message before ending. For the error, well everything is said. – Espy

Answer 1

Error Message:
Exception in thread "main" java.lang.NumberFormatException: For input string: "Ace of Clubs"
    at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)
    at java.lang.Integer.parseInt(Integer.java:580)
    at java.lang.Integer.parseInt(Integer.java:615)
    at set07102.Cards.main(Cards.java:68)
C:\Users\qasim\AppData\Local\NetBeans\Cache\8.1\executor-snippets\run.xml:53: Java returned: 1

means:

There was an error. We try to give you as much information as possible
It was an Exception in main thread. It's called NumberFormatException and has occurred for input "Ace of Clubs".
at line 65th of NumberFormatException.java which is a constructor,
which was invoked from Integer.parseInt() which is in file Integer.java in line 580,
which was invoked from Integer.parseInt() which is in file Integer.java in line 615,
which was invoked from method main in file Cards.java in line 68.

It has resulted in exit code 1

In other words, you tried to parse "Ace of Clubs" to an int what Java can't do with method Integer.parseInt. Java has provided beautiful stacktrace which tells you exactly what the problem is. The tool you're looking for is debugger and using breakpoints will allow you to inspect the state of you application at the chosen moment.

The solution might be the following logic in case you want to use parsing:

if (cards[index].startsWith("Ace")) 
    value = 1;
else if (cards[index].startsWith("King"))
    value = 12;
else if (cards[index].startsWith("Queen"))
    value = 11;
...
else {
    try {
        Integer.parseInt(string.substring(0, cards[index].indexOf(" "))); 
    } catch (NumberFormatException e){
        //something went wrong
    }
}

What is an Exception in Java?

An exception is an event, which occurs during the execution of a program, that disrupts the normal flow of the program's instructions.

-Documentation

Constructors and usage in Integer#parseInt

static NumberFormatException forInputString(String s) {
    return new NumberFormatException("For input string: \"" + s + "\"");
}

public NumberFormatException (String s) {
    super (s);
}

They are important for understanding how to read the stacktrace. Look how the NumberFormatException is thrown from Integer#parseInt:

if (s == null) {
    throw new NumberFormatException("null");
}

or later if the format of the input String s is not parsable:

throw NumberFormatException.forInputString(s); 

What is a NumberFormatException?

Thrown to indicate that the application has attempted to convert a string to one of the numeric types, but that the string does not have the appropriate format.

-Documentation

NumberFormatException extends IllegalArgumentException. It tells us that it's more specialized IllegalArgumentException. Indeed, it's used for highlighting that although, the argument type was correct (String) the content of the String wasn't numeric (a,b,c,d,e,f are considered digits in HEX and are legal when needed).

How do I fix it?
Well, don't fix the fact that it's thrown. It's good that it's thrown. There are some things you need to consider:

1. Can I read the stacktrace?
2. Is the String which causes an Exception a null?
3. Does it look like a number?
4. Is it 'my string' or user's input?
5. to be continued

Ad. 1.

The first line of a message is an information that the Exception occurred and the input String which caused the problem. The String always follows : and is quoted ("some text"). Then you become interested in reading the stacktrace from the end, as the first few lines are usually NumberFormatException's constructor, parsing method etc. Then at the end, there is your method in which you made a bug. It will be pointed out in which file it was called and in which method. Even a line will be attached. You'll see. The example of how to read the stacktrace is above.

Ad. 2.

When you see, that instead of "For input string:" and the input, there is a null (not "null") it means, that you tried to pass the null reference to a number. If you actually want to treat is as 0 or any other number, you might be interested in my another post on StackOverflow. It's available here.

The description of solving unexpected nulls is well described on StackOverflow thread What is a NullPointerException and how can I fix it?.

Ad. 3.

If the String that follows the : and is quoted looks like a number in your opinion, there might be a character which your system don't decode or an unseen white space. Obviously " 6" can't be parsed as well as "123 " can't. It's because of the spaces. But it can occure, that the String will look like "6" but actually it's length will be larger than the number of digits you can see.

In this case I suggest using the debugger or at least System.out.println and print the length of the String you're trying to parse. If it shows more than the number of digits, try passing stringToParse.trim() to the parsing method. If it won't work, copy the whole string after the : and decode it using online decoder. It'll give you codes of all characters.

There is also one case which I have found recently on StackOverflow, that you might see, that the input looks like a number e.g. "1.86" and it only contains those 4 characters but the error still exists. Remember, one can only parse integers with #Integer#parseInt#. For parsing decimal numbers, one should use Double#parseDouble.

Another situation is, when the number has many digits. It might be, that it's too large or too small to fit int or long. You might want to try new BigDecimal(<str>).

Ad. 4.

Finally we come to the place in which we agree, that we can't avoid situations when it's user typing "abc" as a numeric string. Why? Because he can. In a lucky case, it's because he's a tester or simply a geek. In a bad case it's the attacker.

What can I do now? Well, Java gives us try-catch you can do the following:

try {
    i = Integer.parseInt(myString);
} catch (NumberFormatException e) {
    e.printStackTrace();
    //somehow workout the issue with an improper input. It's up to your business logic.
}

Northamptonshire

"Ace of Spades" ??? Also, in your descriptive error detailing, I think you have a typo for line number (615 to 580) for second Integer.parseInt. – Bulbous

When I have time, I'll focus on too big numbers for the point 5. – Northamptonshire

A beautiful answer. I think I should dup-close to this question more often ;-) – Outmoded

@Outmoded That is why I've prepared this answer – Northamptonshire

@OleV.V. sure! Updated – Northamptonshire

Answer 2

What is a NumberFormatException?

This exception is thrown to indicate that the application has attempted to convert a string to one of the numeric types, but that the string does not have the appropriate format.

In your case, according to your stack trace this exception was thrown by Integer.parseInt(String) which means that the provided String does not contain a parseable integer. And still according to the stack trace, it is due to the fact that you tried to parse the String "Ace of Clubs" as an integer which cannot work as it is not the String representation of an integer.

How to fix it?

The simplest and generic way is to catch the exception NumberFormatException

int value = -1;
try {
    value = Integer.parseInt(myString);
} catch (NumberFormatException e) {
    // The format was incorrect
}

It will work but catching an exception is slow because it needs to build the call stack to create the Exception which is costly, so if you can avoid it do it. Moreover you will need to manage the exception properly which is not always obvious.

Or you could use a regular expression to check first if the String matches with an Integer but it is quite error prone as you could easily use a wrong regular expression.

---

In your case, a more OO approach should be used instead of dealing with String, for example you could use a class or an enum to represent your cards instead of using simple String because it is much more error prone as you have already noticed.

So if you decide to use a dedicated class for your card, your code could be:

public class Card {

    private final Rank rank;
    private final Suit suit;

    public Card(final Rank rank, final Suit suit) {
        this.rank = rank;
        this.suit = suit;
    }

    public Rank getRank() {
        return this.rank;
    }

    public Suit getSuit() {
        return this.suit;
    }
}

For the suit and the rank of a card, we can use an enum since there are limited amounts of existing ranks and suits.

public enum Rank {
    ACE(1), TWO(2), THREE(3), FOUR(4), FIVE(5), SIX(6), SEVEN(7), HEIGHT(8),
    NINE(9), TEN(10), JACK(11), QUEEN(12), KING(13);

    private final int value;

    Rank(final int value) {
        this.value = value;
    }

    public int getValue() {
        return this.value;
    }
}

public enum Suit {
    SPADE, HEART, DIAMOND, CLUB
}

Then cards would be an array of Card instead of an array of String, and could be initialized as next:

Rank[] ranks = Rank.values();
Suit[] suits = Suit.values();
Card[] cards = new Card[ranks.length * suits.length];
for (int i = 0; i < ranks.length; i++) {
    for (int j = 0; j < suits.length; j++) {
        cards[i * suits.length + j] = new Card(ranks[i], suits[j]);
    }
}

If you need to shuffle your array of cards, you can proceed as next (please note that if you decide to use a List of cards instead of an array simply use Collections.shuffle(list))

List<Card> allCards = Arrays.asList(cards);
Collections.shuffle(allCards);
allCards.toArray(cards);

Then you will be able to access directly to the value of your card with cards[index].getRank().getValue() without taking the risk to get an exception (except an IndexOutOfBoundsException if you don't use a proper index).

Oshaughnessy

I won't agree with your thesis, that catching NFE is ugly. In large systems, when you can assume, that users will give non numeric input, and on the other hand you want to keep your logs clean, it's better to catch it and log an info or throw your own service exception than leting the whole stacktrace print in red. – Northamptonshire

@Northamptonshire generally speaking catching an exception is slow because it needs to build the call stack to create the Exception which is costly, so if you can avoid it do it but sometimes you just can't avoid it. Here you can avoid it by using a real OO approach which is what I propose – Oshaughnessy

Answer 3

Looks like cards[] is String array and you are trying to convert Ace of Clubs to Integer.

int first_value = Integer.parseInt(cards[index]);

Overtire

Answer 4

java.lang.NumberFormatException 

occurs when you are trying to parse some input which not a Number string.

In your case your trying to parse a string (which not has number )as Integer. As its not possible NumberFormatException exception occured.

int first_value = Integer.parseInt(cards[index]);//cards[index] value should be //number string "123" not "abc"

Canker

Answer 5

A NumberFormatException is the way Java has to say you "I tried to convert a String to int and I could not do it".

In your exception trace you can read

Exception in thread "main" java.lang.NumberFormatException: For input string: "Ace of Clubs"
    at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)
    at java.lang.Integer.parseInt(Integer.java:580)
    at java.lang.Integer.parseInt(Integer.java:615)
    at set07102.Cards.main(Cards.java:68)

Basically, it means that at the line 68 of your code you call to the Integer.parseInt method passing "Ace of Clubs" as paremeter. This method expects a integer value represented as String, e.g. "4", so method complains throwing a NumberFormatException because "Ace of Clubs" does not seem a integer at all.

Arber

Answer 6

A NumberFormatException means that Integer.parseInt() couldn't translate the string into a number.

I would suggest one of two options:

1. Encapsulate cards as a name(string)/value(int) combo. Use the value to do comparisons, and the name to present info to the user. Cards[] then becomes a list of cards, not strings.

2. Parse the strings yourself. Which may be easier, since you've already done it with the if(cards[index].startsWith("Ace")) { value = 1; } bits. You can move those into a function called CardToInt() (or whatever), and use that function instead of Integer.parseInt().

Prickly

Answer 7

The very first thing that threw me for a loop (no pun intended) was you were limiting the value to 1-13 when it needs to be 0-52. Also with your logic the value was always be higher. A better approach is with a number generator. Here is my code using a number generator (or Java Random):

public static void main(String[] args) {

String[] cards = { "Ace of Clubs", "1 of Clubs", "2 of Clubs",
        "3 of Clubs", "4 of Clubs", "5 of Clubs", "6 of Clubs",
        "7 of Clubs", "8 of Clubs", "9 of Clubs", "10 of Clubs",
        "Queen of Clubs", "King of Clubs", "Ace of Diamonds",
        "1 of Diamonds", "2 of Diamonds", "3 of Diamonds",
        "4 of Diamonds", "5 of Diamonds", "6 of Diamonds",
        "7 of Diamonds", "8 of Diamonds", "9 of Diamonds",
        "10 of Diamonds", "Queen of Diamonds", "King of Diamonds",
        "Ace of Hearts", "1 of Hearts", "2 of Hearts", "3 of Hearts",
        "4 of Hearts", "5 of Hearts", "6 of Hearts", "7 of Hearts",
        "8 of Hearts", "9 of Hearts", "10 of Hearts",
        "Queen of Hearts", "King of Hearts", "Ace of Spades",
        "1 of Spades", "2 of Spades", "3 of Spades", "4 of Spades",
        "5 of Spades", "6 of Spades", "7 of Spades", "8 of Spades",
        "9 of Spades", "10 of Spades", "Queen of Spades",
        "King of Spades" };

Scanner scanner = new Scanner(System.in);
Random rand = new Random();
String response = "";
int index = 0;
int value = 0;  
while (!response.equals("q") && index < 52) {

    // set next card value based on current set of cards in play
    if (cards[index].endsWith("Clubs")) {
        value = rand.nextInt(12);
    }
    if (cards[index].endsWith("Diamonds")) {
        value = rand.nextInt(12) + 13;
    }
    if (cards[index].endsWith("Hearts")) {
        value = rand.nextInt(12) + 26;
    }
    if (cards[index].endsWith("Spades")) {
        value = rand.nextInt(12) + 39;
    }

    // display card too user (NOTE: we use the random number not the index)
    System.out.println("Card is: " + cards[value]);

    // ask user what well the next card be
    System.out.println("Will the next card be higher or lower?, press q if you want to quit");
    response = scanner.nextLine();

    // display if user was right (NOTE: compared the random number to the current index)
    // ignore incorrect response and just continue
    if ((value > index && response.startsWith("h")) || (value < index && response.startsWith("l"))) {
        System.out.println("You answer was right, well done!");
    } else {
        System.out.println("You answer was wrong, try again!");
    }

    // continue loop
    index++;
}
}

As for the NumberFormatException I believe Nicolas Filotto did a good job explaining that.

Thenna

Answer 8

The Exception comes in your code, where you convert the String to an Integer :

int first_value = Integer.parseInt(cards[index]);

where you pass a String as "Ace of Clubs" which is not possible to convert as integer,so it throws Number Format Exception. You can use,

try {
     ....
     // Your Code
     ....
    }
catch(NumberFormatException e)
{
    e.getMessage();  //You can use anyone like printStackTrace() ,getMessage() to handle the Exception
}

Coveney

Answer 9

int first_value = Integer.parseInt(cards[index]); 

while writing the above statement, you are trying to parse "Ace of Clubs" as a number.

you can use the following method to test if any string can be parsed as Integer:

boolean tryParseInt(String value) {  
     try {  
         Integer.parseInt(value);  
         return true;  
      } catch (NumberFormatException e) {  
         return false;  
      }  
}

Regarding your question, what is NumberFormatException : It is thrown to indicate that the application has attempted to convert a string to one of the numeric types, but that the string does not have the appropriate format. (ref -http://docs.oracle.com/javase/7/docs/api/java/lang/NumberFormatException.html)

Marga