java - Regex index 0 how it exactly works -

by compiling following:

 system.out.println(pattern.matches(".?(\\d)$","3")); 

it returns true because before 3 there nothing , ? check 1 or zero. 3 first character of input starts @ 0 , end @ 1. how can jvm recognize there nothing before 3. example following.

system.out.println(pattern.matches(".*","hello");

it returns true last character gets matched "nothing". there should not "nothing" character @ beginning of string, @ end of right?

1. this not jvm. java regular expressions.
2. the regular expression ".*" means "match 0 or more characters". it's easy satisfy this, since blank string has 0 characters, , therefore satisfies this. whether java regular expressions choose lazy , match empty string, or greedy , match entire string depends on implementation of java regular expressions. if read excellent writeup (http://docs.oracle.com/javase/tutorial/essential/regex/quant.html) you can see patterns ".*" in java considered "reluctant" quantifiers , prefer take little possible.
3. based on information in writeup, can see pattern ".{0,}" greedy version of same expression. perhaps you'd use instead if problem you.