hex - Java hexadecimal base double literal

Question

I am studying for java certification. And i'm curious about the java literals. I know it is possible to do something like this:

int i = 0xAA;
long l = 0xAAL;

Also this is possible for floating-point variables:

double d = 123d;
float f = 123f;

So I logically thought with these examples that the same would apply for hexadecimal. Just like i can add L for long literals, I could add 'd' or 'f' but the logic is flawed since 'F' and 'D' are valid hexadecimal values.

It is not possible to do something like this:

double d = 0xAAAAAAAAAAAAAAAAAAd;

Is this just not allowed by Java or there is a simple way to do it that I don't know?

question from:https://stackoverflow.com/questions/42722895/java-hexadecimal-base-double-literal

Answer 1

It turns out it is possible, although that surprised me. Section 3.10.2 of the JLS gives the structure of floating point literals, including HexadecimalFloatingPointLiteral.

public class Test {

    public static void main(String[] args) {
        double d1 = 0xAAAAAAAAAAAAAAAAAAp0d;
        double d2 = 0x1.8p1d;

        System.out.println(d1); // A very big number
        System.out.println(d2); // 24 = 1.5 * 2^1
    }
}

The p is required as part of the binary exponent - the value after the p is the number of bits to shift the value left. Other examples:

0x1.4p0d => 1.25 (binary 0.01 shifted 0 bits)
0x8p-4d => 0.5 (binary 1000 shifted *right* 4 bits)