Array.sort() question

Question

I've been looking at Arrays and the Comparator interface. But I don't understand the sorting results:
import java.util.*;
import java.io.*;
class SearchObjArray {
     public static void main(String[] args) {
          Console con=System.console();
          if (con==null) System.exit(1);
          String[] sa={"one","two","three","four"};
          Arrays.sort(sa);          
          con.format("The sorted string:%n");          
          for(String s: sa)
               con.format("%s ",s);
          //four,one,three,two
          con.format("%nA binary search for \"one\": one=%s%n",Arrays.binarySearch(sa,"one"));
          //one=1, as expected. The binarySearch found "one" in position 1 of the sorted array.
          ReSortComparator rs=new ReSortComparator();     
          con.format("A binary search for \"one\" using the resort comparator on the string array: one=%s%n",Arrays.binarySearch(sa,"one",rs));
          //one=-1<====WHY -1? Why is the insertion point 0?
     static class ReSortComparator implements Comparator<String> {
          public int compare(String a, String b) {
               return b.compareTo(a);
          

Answer 1

You have to search using the same ordering rules as how the array is sorted. You sorted with one set of rules (String's natural ordering) and searched with a different set of rules (the opposite ordering, as determined by your comparator). That's the same as trying to do a binary search on an unsorted array. Since one of the preconditions of binary search is that the array be sorted, it shouldn't be surprising that you get odd results.