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Home/ Questions/Q 204747
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Editorial Team
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Editorial Team
Asked: May 11, 20262026-05-11T17:29:42+00:00 2026-05-11T17:29:42+00:00

public class Main3 { public static void main(String[] args) { Integer min = Integer.MIN_VALUE;

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public class Main3 {
    public static void main(String[] args) {
        Integer min = Integer.MIN_VALUE;
        String minHex = Integer.toHexString(Integer.MIN_VALUE);

        System.out.println(min + " " + minHex);
        System.out.println(Integer.parseInt(minHex, 16));
    }
}

Gives

-2147483648 80000000
Exception in thread "main" java.lang.NumberFormatException: For input string: "80000000"
    at java.lang.NumberFormatException.forInputString(NumberFormatException.java:48)
    at java.lang.Integer.parseInt(Integer.java:459)
    at Main3.main(Main3.java:7)

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-11T17:29:43+00:00Added an answer on May 11, 2026 at 5:29 pm

    It’s documented that Integer.toHexString returns a string representation of the integer as an unsigned value – while Integer.parseInt takes a signed int. If you use Integer.toString(value, 16) instead you’ll get what you want.

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