重写hasHashCode时如何计算对象的HashCode

From Effective Java Sencond Edition Item9

1. Store some constant nonzero 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, that is), 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 com-                
       pares 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. When you are finished writing the hashCode method, ask yourself whether equal instances have equal hash codes. Write unit tests to verify your intuition! If equal instances have unequal hash codes, figure out why and fix the problem.