How to use .valueOf function in java without losing precision?

Question

I want to convert a string to its integer value in java so I'm using the .ValueOf function, however, it truncates the beginning of the string when making it an integer.

For example, converting the string "0900" to the integer value 0900 by using:

int x = Integer.valueOf("0900");

How can I get x to be 0900 instead of 900?

There's no such distinction in the representation of an integer. In both cases it is a word of exactly 32 bits, with value 0000 0000 0000 0000 0000 0011 1000 0100 — undefined symbol, Mar 11 '23 at 19:12
An int is just a quantity. It has no concept of leading zeroes, just as you would not distinguish between holding 3 apples and 03 apples. If you want to *show* a leading zero, you can use `String.format("%04d", x)`. — VGR, Mar 11 '23 at 19:31
Or, for that matter, "three apples". It's just another representation of the number 3 as applied to apple quantity. — undefined symbol, Mar 11 '23 at 21:51

score 0 · Answer 1 · answered Mar 12 '23 at 12:51

if you have a special use-case where you need to keep this "0" prefix as well as use it as a numeric value, you should build a custom class for it. For instance, let's assume this number is introduced by the user in some input field on the screen. the class can look like this (it can even be a record):

public class UserInput {
    private final String originalString;
    private final int value;

    public UserInput(String originalString) {
        this.originalString = originalString;
        this.value = Integer.valueOf(originalString);
    }

    public String text() {
        return originalString;
    }

    public int numericValue() {
        return value;
    }
}

now you can create UserInput by providing the original Stirng and use any of the numericValue method for comparison and mathematical operations:

UserInput previousInput = new UserInput("000039");
UserInput input = new UserInput("0040");

System.out.println("value introduced: " + input.text());
System.out.println("numeric value increased: " + input.numericValue() > previousInput.numericValue());

is this what you needed?

How to use .valueOf function in java without losing precision?

1 Answers1