Help with Greedy and Reluctant SCPJ 310 055

[RAUL MORALES]

Hi friends I have some misgivings with greedy and reluctant
in regex expersions . I ran this piece of code

import java.util.regex.*;
public class Music{
String s = " rap rapture wrap" ;
public static void main(String argv[]){
new Music();
}
Music(){
String regex = "r\\wp?";
Pattern pattern = Pattern.compile(regex);
Matcher matcher =pattern.matcher(s);
while(matcher.find()){
System.out.print(matcher.group());
}
}
}

and the output was rapraprerap with the greedy ? that means 0 o 1 character is reading from
left to right but the book say that greedy read all the string and begin from the end to the
begining.After that I modified the code with the reluctant operator as below


import java.util.regex.*;
public class Music{
String s = " rap rapture wrap" ;
public static void main(String argv[]){
new Music();
}
Music(){
String regex = "r\\wp??";
Pattern pattern = Pattern.compile(regex);
Matcher matcher =pattern.matcher(s);
while(matcher.find()){
System.out.print(matcher.group());
}
}
}

and the output was rararera . I am confused somebody can help me.

[Henry Wong]

I am confused somebody can help me.

Quite frankly, I am not sure what you are asking here, as you don't seem to say what you are confused about... However...

There are two behaviors you need to keep track of. First, yes, in a greedy match, the regex will try to match as much as possible. And as the matches fail, it will back off until the match succeeds.

Second, and this is not related to greedy or reluctant, is how the find() method behaves. The find() method will try to match starting from the end of the last match, going forward, until it succeeds. And this is the behavior regardless of whether the regex is greedy or reluctant.

Henry