Integer printing wrong value

Question

When converting an integer to int array, for example 123 to {1,2,3}, I am getting values {49,50,51}. Not able to find what is wrong with my code.

public class Test {
    public static void main(String [] args) {
        String temp = Integer.toString(123);
        int[] newGuess = new int[temp.length()];
        for (int i = 0; i < temp.length(); i++) {
            newGuess[i] = temp.charAt(i);
        }
        for (int i : newGuess) {
            System.out.println(i);
        }
    }
}

Output:

49

50

51

Possible duplicate of [Converting an integer into an int array in Java](http://stackoverflow.com/questions/39482457/converting-an-integer-into-an-int-array-in-java) — jitendra varshney, Sep 14 '16 at 05:41

score 4 · Accepted Answer · edited Jun 20 '20 at 09:12

4

charAt(i) will give you UTF-16 code unit value of the integer for example in your case, UTF-16 code unit value of 1 is 49. To get integer representation of the value, you can subtract '0'(UTF-16 code unit value 48) from i.

public class Test {
    public static void main(String [] args) {
        String temp = Integer.toString(123);
        int[] newGuess = new int[temp.length()];
        for (int i = 0; i < temp.length(); i++) {
            newGuess[i] = temp.charAt(i);
        }
        for (int i : newGuess) {
            System.out.println(i - '0');
        }
    }
}

Output:

1

2

3

edited Jun 20 '20 at 09:12

Community

1
1

answered Sep 14 '16 at 05:14

codingenious

8,385
12
60
90

1

To be pedantic, `charAt` doesn't return ASCII values, but rather `char` values, which are UTF-16 code units. It just so happens that the first 128 characters are the same either way, which covers the common Latin alphabet. But Java has never used ASCII strings, it's been Unicode from day one. – Daniel Pryden Sep 14 '16 at 06:35

score 2 · Answer 2 · answered Sep 14 '16 at 06:17

To add a little Java 8 niceties to the mix which allows us to pack everything up neatly, you can optionally do:

int i = 123;
int[] nums = Arrays.stream(String.valueOf(i).split(""))
        .mapToInt(Integer::parseInt)
        .toArray();

Here we get a stream to an array of strings created by splitting the string value of the given integer's numbers. We then map those into integer values with Integer#parseInt into an IntStream and then finally make that into an array.

score 1 · Answer 3 · answered Sep 14 '16 at 05:22

temp.charAt(i) is basically returning you characters. You need to extract the Integer value out of it.

You can use:

newGuess[i] = Character.getNumericValue(temp.charAt(i));

Output

1
2
3

Code

public class Test {
    public static void main(String [] args) {
        String temp = Integer.toString(123);
        int[] newGuess = new int[temp.length()];
        for (int i = 0; i < temp.length(); i++) {
            newGuess[i] = Character.getNumericValue(temp.charAt(i));
        }
        for (int i : newGuess) {
            System.out.println(i);
        }
    }
}

score 0 · Answer 4 · answered Sep 14 '16 at 05:25

0

As your interest is to get as an integer value of an string. Use the method parse int Integer.parseInt() . this will return as integer. Example : int x =Integer.parseInt("6"); It will return integer 6.

answered Sep 14 '16 at 05:25

sanjaya Kumar panigrahy

121
8

`Integer.parseInt("123")` will return integer `123`, not `{1,2,3}` int array. – codingenious Sep 14 '16 at 05:29

Integer printing wrong value

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