Question

Suppose a class MyClass has 3 variables, a,b and c. This class also overrides hashCode() that calculates return value using Joshua Bloch’s approach.

Assuming the above, would it be correct to assume that MyClass() and new MyClass() are deeply equal if their hashcodes match?

For the purposes of the example, assume that both objects are initialized with the same parameters

Answer 1

Certainly not.

hashCode is required to yield the same value for equivalent objects, but different objects may yield the same value. So, the following is a correct (but inefficient) hashCode implementation:

public int hashCode()
{
    return 42;
}

Even when you know a specific hashing method is used, it is a bad idea to assume hashCode is perfect: later, some refactoring/subclass might replace the hashCode implementation with something else. If you want this, create a perfectHashCode method that does always return a perfect hashcode, and use that for implementing hashCode

To answer your initial question: no, even Bloch’s approach afaics does not yield a perfect hashCode, because (because of wrapping over MAXINT) 2 different paths might eventually collide into the same hash code.

This is easy to imagine: say your object holds 2 integers. The hash code is only 1 integer. So it can never yield a different value for each combination of 2 integers.