Java error when converting String in ArrayList to int

Question

I am very new at Java. I am trying to write a lottery simulator. I have the following code:

Scanner input = new Scanner(System.in);
System.out.println("Enter six whole numbers between 1 and 49 in a comma-separated list: ");
String userNumbersString = input.nextLine();
String[] userNumbersArray = userNumbersString.replaceAll("\\s+","").split(",");
ArrayList userNumbers = new ArrayList(Arrays.asList(userNumbersArray));
boolean validNumbers = true;
for(int v = 0; v <= 6; v++){
    if((int)userNumbers.get(v) > 49){
        validNumbers = false;
            v = 7;
        }else{
            v++;
        }
    }
if(validNumbers == false){
    String[] str = new String[0];
    main(str);
}

It compiles properly. However, when I input a series of numbers, I get this error:

java.lang.ClassCastException: java.lang.String cannot be cast to java.lang.Integer.

I am very confused about why this string can't be cast to an integer.

Sorry if this is a long-winded question, but I would hugely appreciate any help.

I have just realised that the OutOfBounds exception was due to a stupid maths mistake on my part; v was starting at 0; there were five values in the array, but I gave the for loop the condition <= 6. Sorry about that.

Thank you everybody for all your help. Together, you have resolved my question.

Don't include massages for one person in question itself. Leave it as comment under this person answer. — Pshemo
– Pshemo, Commented Sep 20, 2014 at 20:55

nem035 · Accepted Answer · 2014-09-20 21:04:47Z

2

In simple terms

int is a primitive type for representing a 32-bit integer number. String is a reference type wrapped arround an array of chars to represent a sequence of characters.

You cannot just change the type from int to String and make the character sequence in the string magically becomes a number.

Think about how you would implement this conversion. You would look at each character, make sure it is a valid digit etc.

Change:

if((int)userNumbers.get(v) > 49)

To:

if(Integer.parseInt(userNumbers.get(v)) > 49)

Further, when creating Generic type containers like ArrayList make sure you also declare what type of objects they store for type safety reasons.

Change:

ArrayList userNumbers = new ArrayList(Arrays.asList(userNumbersArray));

To:

ArrayList<String> userNumbers = new ArrayList<String>(Arrays.asList(userNumbersArray));

This way you calling userNumber.get() will return a String. If you don't put this it will return an Object so you would actually call Integer.parseInt(Object) which is invalid. You would get a pretty self-explanatory error looking like this:

The method parseInt(String) in the type Integer is not applicable for the arguments (Object)

EDIT:

For your IndexOutOfBounds error, you are trying to access more elements than you have. Basically, you want to have 6 values but your for loop says from 0 until 6 (included). That is a total of 7 elements: 0, 1, 2, 3, 4, 5, 6

You can fix this by changing the <= 6 to < 6 in your loop but the best way to fix it so it will work no matter how many elements you have is to use the size() property of ArrayList:

Change:

for(int v = 0; v <= 6; v++)

To:

for(int v = 0; v < userNumbers.size(); v++)

edited Sep 20, 2014 at 21:04

answered Sep 20, 2014 at 20:44

nem035

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7 Comments

user3315726 Over a year ago

Thanks for the prompt reply but, when I do this, I get a compilation error. I will update my original question with the error as it is quite long.

bcsb1001 Over a year ago

Why valueOf instead of parseInt? In this case, there is no reason to use an Integer instead of int.

nem035 Over a year ago

@bcsb1001 you are correct, although it still works, parseInt is a better choice. Updated my answer

Pshemo Over a year ago

@user3315726 valueOf expects String as argument, while get returns Object because you didn't specified what type of elements should ArrayList store. To do this you would have to use generic types (also prefer interfaces over implementation) so change ArrayList userNumbers = new ArrayList(... to List<String> userNumbers = new ArrayList<String>(....

nem035 Over a year ago

@Pshemo haha I literally just added that part, telepathy wow :) thanks for the suggestion

|

bcsb1001 · Accepted Answer · 2014-09-20 20:46:56Z

0

First, you should use generics for ArrayList. Please replace

ArrayList userNumbers = new ArrayList(Arrays.asList(userNumbersArray));

with

ArrayList<String> userNumbers = new ArrayList<>(Arrays.asList(userNumbersArray));

However, this is not your problem. The misconception here is that casting will infer what type of conversion you need; it won't. Replace (int)userNumbers.get(v) with Integer.parseInt(userNumbers.get(v)) to parse the int from a String.

answered Sep 20, 2014 at 20:46

bcsb1001

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1 Comment

user3315726 Over a year ago

I have now used generics and it will compile with parseInt, but I get an error after inputting the comma-separated number string: java.lang.IndexOutOfBoundsException: index: 6, size: 6 (in java.util.ArrayList)

dfherr · Accepted Answer · 2014-09-20 20:49:46Z

0

Type casting is not meant to parse an Integervalue from an String.

You should use Integer.parseInt for that.

if(Integer.parseInt(userNumbers.get(v)) > 49

should solve your error.

answered Sep 20, 2014 at 20:49

dfherr

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Comments

Michael Krause · Accepted Answer · 2014-09-21 07:24:30Z

First, you will get an ArrayIndexOutOfBounds exception because of the <= 6 in your for loop.

Secondly, you should use the Integer.parseInt(String intStr) method to turn each number into its int equivalent.

Here is a stripped down version that I think works:

public static void main(String[] args)
{
    String testInputOne = "14,22,18,2,5,11";

    boolean allNumbersAreValid = true;

    String[] userNumbersStrArray = testInputOne.split(",");

    for (String numberStr: userNumbersStrArray)
    {
        Integer number = Integer.parseInt(numberStr.trim());

        if (number > 49)
        {
            allNumbersAreValid = false;
            break;
        }
    }

    System.out.println("Testing " + testInputOne + ": " + (allNumbersAreValid?"All valid!":"At least one invalid"));
}

If you want to further guard against bad input such as someone entering an alpha character, you could wrap the Integer.parseInt call in a try catch block and catch the NumberFormatException and output a suitable error message.

I'm frankly a little scared as to what the recursive call to main is supposed to do? :)

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