Question

This simple regex program

import java.util.regex.*;
class Regex {
    public static void main(String [] args) {
        System.out.println(args[0]); // #1
        Pattern p = Pattern.compile(args[0]); // #2
        Matcher m = p.matcher(args[1]);
        boolean b = false;
        while(b = m.find()) {
            System.out.println(m.start()+" "+m.group());
        }
    }
}

invoked by java regex "\d" "sfdd1" compiles and runs fine.

But if #1 is replaced by Pattern p = Pattern.compile("\d");, it gives compiler error saying illegal escape character. In #1 I also tried printing the pattern specified in the command line arguments. It prints \d, which means it is just getting replaced by \d in #2.

So then why won’t it throw any exception? At the end it’s string argument that Pattern.compile() is taking, doesn’t it detect illegal escape character then? Can someone please explain why is this behaviour?

Answer 1

A backslash character in a string literal needs to be escaped (preceded by a backslash). When passed in from the command line the string is not a string literal. The compiler complains because "\d" is not a valid escape sequence (see Escape Sequences for Character and String Literals ).