Casting string into <Integer>ArrayList

Question

    Scanner scan = new Scanner(System.in);
    System.out.println("Enter a sequence of numbers ending with 0.");

    ArrayList<Integer> list = new ArrayList<Integer>();

    String num = scan.nextLine();

    for(int x=0; x < num.length(); x++){
        System.out.println(num.charAt(x));

        int y = num.charAt(x);
        System.out.println(y);
        list.add(y);
        System.out.println(list);


    }

Im trying to cast a string of numbers into a array. Its not adding the correct vaule. I keep getting 49 and 50. I want to store the numbers the user enters into the ArrayList. Can someone help?

That is because its giving you the ASCII value, int y = num.charAt(x)-48 or Character.valueOf(num.charAt(x)) since '0' is represented by 48, refer: asciitable.com — Nitin Dandriyal
– Nitin Dandriyal, Commented May 12, 2015 at 4:59

Thilo · Accepted Answer · 2015-05-12 04:50:04Z

2

 int y = num.charAt(x);

That will give you the Unicode codepoint for the character. Like 65 for A or 48 for 0.

You probablay want

 int y = Integer.parseInt(num.substring(x, x+1));

answered May 12, 2015 at 4:50

Thilo

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Comments

Rahul Tripathi · Accepted Answer · 2015-05-12 04:50:26Z

0

You can try to use:

int y = Integer.parseInt(num.charAt(x));

instead of

int y = num.charAt(x);

answered May 12, 2015 at 4:50

Rahul Tripathi

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1 Comment

Naman Gala Over a year ago

Integer.parseInt is not applicable for char as a method argument. You code will give compilation error.

Saurabh Jhunjhunwala · Accepted Answer · 2015-05-12 04:54:10Z

0

You are not converting the input to Integer, thus JVM is taking them as string. Assuming you are 1 as you input it is printing 49 (ASCII equivalent) of "1".

If you want to get the integral values, you need to parse it using

int y = Integer.parseInt(num.charAt(x));
System.out.println(y);
list.add(y);
System.out.println(list);

answered May 12, 2015 at 4:54

Saurabh Jhunjhunwala

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1 Comment

Naman Gala Over a year ago

Integer.parseInt is not applicable for char as a method argument. You code will give compilation error.

Naman Gala · Accepted Answer · 2015-05-12 05:27:58Z

As this code int y = num.charAt(x); is creating the problem. As you are trying to store returned character into int value, so it is storing ASCII value of the character.

You can go with the suggestions in other answers.

For the simplicity, you can rewrite your code like this.

Scanner scan = new Scanner(System.in);
System.out.println("Enter a sequence of numbers ending with 0.");

ArrayList<Integer> list = new ArrayList<Integer>();

String num = scan.nextLine();

char[] charArray = num.toCharArray();
for (char c : charArray) {
    if (Character.isDigit(c)) {
        int y = Character.getNumericValue(c);
        System.out.println(y);
        list.add(y);
        System.out.println(list);
    } else {
         // you can throw exception or avoid this value.
    }
}

Note: Integer.valueOf and Integer.parseInt will not give proper result for char as a method argument. You need to pass String as a method argument in both the cases.

techPackets · Accepted Answer · 2015-05-12 06:08:22Z

0

You are copying a char into an int. You need to convert it into an int value.

int y = Character.getNumericValue(num.charAt(x));

edited May 12, 2015 at 6:08

answered May 12, 2015 at 4:57

techPackets

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5 Comments

Naman Gala Over a year ago

Your method is taking integer as an input (Integer valueOf(int i)), so in this case it will return ASCII value.

techPackets Over a year ago

check this out docs.oracle.com/javase/7/docs/api/java/lang/… besides num.charAt(x) does not return an int.

Naman Gala Over a year ago

That is for String, not for char. In your code, it is passing char, so internally it will call valueOf(int i) method, and it will pass ASCII value of char (returned by num.charAt(x)) as a method argument. You can check by executing your code.

techPackets Over a year ago

You are right, valueOf method doesn't give the correct value;

Naman Gala Over a year ago

Your code can work like this int y = Integer.valueOf(String.valueOf(num.charAt(x)));. Refer note part of my answer to this question.

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