Today, I felt on a really nice blog post about how a switch is implemented for String under the hood.

It basically explains that the switch is implemented on the String hash code. Easy and efficient. But it made me wondering what would happen if the was a collision in the hash codes. By that I mean: What happen if you got two cases containing strings having the same hash code?

To test it, I've created such a switch. First I was needing to found two strings with the same hash code. The way the hash code of calculate for a string is quite simple. It loops over the characters, adding their value to the previous result multiplied by 31. So, doing a bit of algebra, the two characters strings below will be in collision.

int c1 = some_character, c2 = some_other_character;
int c3 = c1-1, c4 = 31*c1+c2-31*c3;

String s1 = new String(new char[] {(char)c1, (char)c2});
String s2 = new String(new char[] {(char)c3, (char)c4});

From where I just had to create a switch with the result. For instance "10" and "0O" both have a hash code of 1567.

switch (args[0]) {
case "10":
    System.out.println("Here");
    break;
case "0O":
    System.out.println("There");
    break;
case "123":
    System.out.println("Away");
    break;
}

Decompiling this code brings a surprising result.

String str1 = args[0];int i = -1;
switch (str1.hashCode())
{
case 1567:
  if (str1.equals("0O")) {
    i = 1;
  } else if (str1.equals("10")) {
    i = 0;
  }
  break;
case 48690:
  if (str1.equals("123")) {
    i = 2;
  }
  break;
}
switch (i)
{
case 0:
  System.out.println("Here");
  break;
case 1:
  System.out.println("There");
  break;
case 2:
  System.out.println("Away");
}