Java negative int to hex and back fails

Question

Hello,

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)

Whats up?

Answer 1

You need to include a negative sign.

I don't have access to test this right now but I'd bet if you tried this value instead:

Integer min = Integer.MIN_VALUE + 1;

It wouldn't bomb, but would give you a positive number (not negative) when you ran ParseInt(min,16).

A string of bits doesn't really have enough info to determine sign in this context so you need to provide it. (consider the case where you use min = "F". Is that +/-F? If you converted it to bits and saw 1111, and you knew it was a byte, you might conclude that it's negative, but that's a lot of ifs.

Answer 2

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.

Answer 3

Try this:

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));
    }

}

to get this:

-2147483648 80000000
-2147483648

Answer 4

This is something that's always annoyed me. If you initialize an int with a hex literal, you can use the full range of positive values up to 0xFFFFFF; anything larger than 0x7FFFFF will really be a negative value. This is very handy for bit masking and other operations where you only care about the locations of the bits, not their meanings.

But if you use Integer.parseInt() to convert a string to an integer, anything larger than "0x7FFFFFFF" is treated as an error. There's probably a good reason why they did it that way, but it's still frustrating.

The simplest workaround is to use Long.parseLong() instead, then cast the result to int.

int n = (int)Long.parseLong(s, 16);

Of course, you should only do that if you're sure the number is going to be in the range Integer.MIN_VALUE..Integer.MAX_VALUE.