A good hash function tends to produce unequal hash codes for unequal objects. Ideally, a hash function should distribute any reasonable collection of unequal instances uniformly across all possible hash values. Achieving this would be ideal, but can be complex. Luckily the following rules made this easy:

1. Store some constant non zero value, say, 17, in an int variable called result.

2. For each significant field f in your object (each field taken into account by the

equals method), do the following:

a. Compute an int hash code c for the field:

i. If the field is a boolean, compute (f ? 1 : 0).

ii. If the field is a byte, char, short, or int, compute (int) f.

iii. If the field is a long, compute (int) (f ^ (f >>> 32)).

iv. If the field is a float, compute Float.floatToIntBits(f).

v. If the field is a double, compute Double.doubleToLongBits(f), and then hash the resulting long as in step 2.a.iii.

vi. If the field is an object reference and this class’s equals method, compares the field by recursively invoking equals, recursively invoke hashCode on the field. If a more complex comparison is required, compute a “canonical representation” for this field and invoke hashCode on the canonical representation. If the value of the field is null, return 0 (or some other constant, but 0 is traditional).

vii. If the field is an array, treat it as if each element were a separate field. That is, compute a hash code for each significant element by applying these rules recursively, and combine these values per step 2.b. If every element in an array field is significant, you can use one of the Arrays.hashCode methods added in release 1.5.

b. Combine the hash code c computed in step 2.a into result as follows:

result = 31 * result + c;

3. Return result.

4. Write unit tests to verify if equal instances have unequal hash codes, figure out why and fix the problem.

1. Store some constant non zero value, say, 17, in an int variable called result.

2. For each significant field f in your object (each field taken into account by the

equals method), do the following:

a. Compute an int hash code c for the field:

i. If the field is a boolean, compute (f ? 1 : 0).

ii. If the field is a byte, char, short, or int, compute (int) f.

iii. If the field is a long, compute (int) (f ^ (f >>> 32)).

iv. If the field is a float, compute Float.floatToIntBits(f).

v. If the field is a double, compute Double.doubleToLongBits(f), and then hash the resulting long as in step 2.a.iii.

vi. If the field is an object reference and this class’s equals method, compares the field by recursively invoking equals, recursively invoke hashCode on the field. If a more complex comparison is required, compute a “canonical representation” for this field and invoke hashCode on the canonical representation. If the value of the field is null, return 0 (or some other constant, but 0 is traditional).

vii. If the field is an array, treat it as if each element were a separate field. That is, compute a hash code for each significant element by applying these rules recursively, and combine these values per step 2.b. If every element in an array field is significant, you can use one of the Arrays.hashCode methods added in release 1.5.

b. Combine the hash code c computed in step 2.a into result as follows:

result = 31 * result + c;

3. Return result.

4. Write unit tests to verify if equal instances have unequal hash codes, figure out why and fix the problem.