I want to create a an Array with Hex pointers!!

Question

Hi all,
I want to create an array that uses Hex as addresses, so that if I want to call on the 16th array element I wouldn't use array[15] but would use array[F] instead, and if I want the 18th element it would be array[11]. Is there any good way of doing that?

Answer 1

They're just numbers. You increment them by adding one to them...
    int x = 0x43AF;
    int y = 232 + x;
    int z = 0x9A + 0x10B + 4 + 5 - 0x88FEF;
    int a = x + y;
    z ++;
    a += y;
    x += (0x9F9F - 55 * y);They could be bytes, too, or they could be longs, shorts, or chars. It's just a different way to type constants. The way you create your array isn't any different than normal, nor is the way you treat the numbers. The primitive data types that can support those numbers doesn't change either... they're just integers, but with a different notation.
As far as converting an integer to a hexadecimal String, you can use the appropriate wrapper object's toHexString method:
    int x = 0xBEEF; // doesn't matter if you type 0xBEEF or 48879 or 0137357
    System.out.println("x = " + x + " decimal");
    System.out.println("x = " + Integer.toHexString(x) + " hex");
    System.out.println("x = " + Integer.toBinaryString(x) + " binary");
    System.out.println("x = " + Integer.toOctalString(x) + " octal");If you want to parse a hexadecimal string that the user enters, use the parseXXX method in the appropriate wrapper class. Keep in mind that, for hexadecimal numbers, these methods will get confused by a "0x" at the beginning. So, if you are converting a String to an integer using base 16, you'll have to make sure you strip off the "0x" at the beginning if there is one.
    int x = Integer.parseInt("BEEF", 16);
    int y = Integer.parseInt("48879", 10); // '10' is optional here
    int z = Integer.parseInt("137357", 8);
    int a = Integer.parseInt("1011111011101111", 2);x, y, z, and a should all end up with the same value if I did the conversions right.
Go to http://java.sun.com/docs/books/jls/second_edition/html/lexical.doc.html#48298 and read section 3.10.
Jason