why is this a good implementation of `hashCode()`

Question

I saw a code that overrides equals() and understandably override hashcode() correspondingly.

 @Override
 public int hashCode() {
 int result = id;
 result = 31 * result + (int) (providerId ^ (providerId >>> 32));
 result = 31 * result + (int) (promotionId ^ (promotionId >>> 32));
 result = 31 * result + (coordinate != null ? coordinate.hashCode() : 0);
 return result;
 }

Can someone please explain why is this hashCode good? Why is it unique to a specific object?

What does >> vs. >>> means? I thought the >> do x bit shifts to the right which is like multiplying by 2.

But what does >>> mean?

• The Paramagnetic Croissant: How do you know it's good?
• yshavit: Note that hash functions do not have to be unique per equal object. It must be the case that equal objects have equal hashcodes, but the contrapositive is not true: unequal objects do not have to have unequal hash codes. Consider strings, for example. There are a theoretically infinite number of strings ("a", "aa", "aaa"...), yet only 2^32 different integer values. Since infinity is larger than 2^32, there must be some strings that are unequal yet have the same int hash code.

Answer 1

Can someone please explain why is this hashCode good? Why is it unique to a specific object?

It's not; essentially no hash code function is. It's just supposed to be rare that two objects have the same hash code, not impossible.

What does >> vs. >>> means? I thought the >> do x bit shifts to the right which is like multiplying by 2.

But what does >>> mean?

It's an unsigned right shift. >> shifts in 1s to the left if the first bit was already 1, >>> shifts in 0s no matter what the original bits were.

• Elad Benda: but is this pattern make this hash rare? I try to understand why? how the bits usually look like. Or I shouldn't think that way?
• Louis Wasserman: The point of hash functions is to mess the bits around as much as possible, the messier, the better. Don't try to work out how the bits in the result look.