Java Regex for matching hexadecimal numbers in a file

Question

So I'm reading in a file (like java program < trace.dat) which looks something like this:

58
68
58
68
40
c
40
48
FA

If I'm lucky but more often it has several whitespace characters before and after each line.

These are hexadecimal addresses that I'm parsing and I basically need to make sure that I can get the line using a scanner, buffered reader... whatever and make sure I can then convert the hexadecimal to an integer. This is what I have so far:

Scanner scanner = new Scanner(System.in);
int address;
String binary;
Pattern pattern = Pattern.compile("^\\s*[0-9A-Fa-f]*\\s*$", Pattern.CASE_INSENSITIVE);
while(scanner.hasNextLine()) {
    address = Integer.parseInt(scanner.next(pattern), 16);
    binary = Integer.toBinaryString(address);
    //Do lots of other stuff here
}
//DO MORE STUFF HERE...

So I've traced all my errors to parsing input and stuff so I guess I'm just trying to figure out what regex or approach I need to get this working the way I want.

Answer 1

The s.next() takes care of the white-spaces. (The default tokenizer doesn't care about them.)

import java.util.Scanner;
public class Test {
    public static void main(String... args) {
        Scanner s = new Scanner(System.in);
        while (s.hasNext())
            System.out.println(Integer.parseInt(s.next(), 16));
    }
}

If you'd really like to stick with the Pattern-approach, I would recommend you to use the XDigit class:

\p{XDigit} A hexadecimal digit: [0-9a-fA-F]

Further more; The scanner.next(pattern) will return the entire matched pattern (including the white-spaces!) You need to work with capturing groups. Try the pattern

^\\s*(\\p{XDigit}+)\\s*$

And then get the actual hex-number with matcher.group(1)