How to replace regex match into a char value (in the middle of a string)

Question

Hi uncle_alice and other great regex gurus
One of my friends has a peculiar problem and I cant give him a solution.
Using String#replaceAll(), i.e. NOT a Matcher loop, how could we convert matched digit string such as "65" into a char of its numeric value. That is, "65" should be converted into letter 'A'.
Here's the failing code:
public class GetChar{
  public static void main(String[] args){
    String orig = "this is an LF<#10#> and this is an 'A'<#65#>";
    String regx = "(<#)(\\d+)#>";
    //expected result : "this is an LF\n and this is an 'A'A"
    String result = orig.replaceAll(regx, "\\u00$2");
    // String result = orig.replaceAll(regx, "\\\\u00$2"); //this also doesn't work
    System.out.println(result);

Answer 1

I don't know that we have lost anything substantial.i think its just that the kind of task this is
especially useful for is kind of a blind-spot in the
range of things java is a good-fit for (?)
for certain tasks (eg process output munging) an
experienced perl programmer could knock up (in perl)
using built-in language features a couple of lines
which in java could takes pages to do. If the cost is
readability/maintainability/expandability etc.. then
this might be a problem, but for a number of
day-to-day tasks it isn't
i'm trying to learn perl at the moment for this exact
reason :)Yes. And when a Java source-code processor(a.k.a. compiler) sees the code like:
line = line.replaceAll(regexp,  new String(new char[] {(char)(Integer.parseInt("$1"))}));or,
line = line.replaceAll(regexp,  doMyProcessOn("$1")); //doMyProcess returns a Stringa common sense should have told him that "$1" isn't a literal string "$1" in this regular expression context.
By the way, I abhor Perl code becaus of its incomprehensibleness. They can't be read by an average common sense. Java code can be, sort of ...