By "non-empty", I mean in this question a string which contains at least one non-zero character.
For reference, here's the hashCode
implementation :
1493 public int hashCode() {
1494 int h = hash;
1495 if (h == 0) {
1496 int off = offset;
1497 char val[] = value;
1498 int len = count;
1499
1500 for (int i = 0; i < len; i++) {
1501 h = 31*h + val[off++];
1502 }
1503 hash = h;
1504 }
1505 return h;
1506 }
and the algorithm is specified in the documentation.
Before an integer overflow occurs, the answer is easy: it's no. But what I'd like to know is if, due to integer overflow, it's possible for a non-empty string to have a hashcode of zero? Can you construct one?
What I'm looking for would ideally be a mathematical demonstration (or a link to one) or a construction algorithm.
Sure. The string f5a5a608 for example has a hashcode of zero.
I found that through a simple brute force search:
public static void main(String[] args){
long i = 0;
loop: while(true){
String s = Long.toHexString(i);
if(s.hashCode() == 0){
System.out.println("Found: '"+s+"'");
break loop;
}
if(i % 1000000==0){
System.out.println("checked: "+i);
}
i++;
}
}
Edit: Joseph Darcy, who worked on the JVM, even wrote a program that can construct a string with a given hashcode (to test the implementation of Strings in switch/case statements) by basically running the hash algorithm in reverse.
just be care of that int h;
. It may overflow, every string that satisfy h % 2^31 == 0
may lead to this.
public class HelloWorld {
public static void main(String []args) {
System.out.println("\u0001!qbygvW".hashCode());
System.out.println("9 $Ql(0".hashCode());
System.out.println(" #t(}lrl".hashCode());
System.out.println(" !!#jbw}a".hashCode());
System.out.println(" !!#jbw|||".hashCode());
System.out.println(" !!!!Se|aaJ".hashCode());
System.out.println(" !!!!\"xurlls".hashCode());
}
}
A lot of strings...