Effective Java 09 Always override hashCode when you override equals

Failure to do so will result in a violation of the general contract for Object.hashCode, which will prevent your class from functioning properly in conjunction with all hash-based collections, including HashMap, HashSet, and Hashtable.

Simple recipe for override hashCode

1. Store some constant nonzero value, say, 17, in an invariable called result.

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

equalsmethod, that is), do the following:

a. Compute an inthash code cfor 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.

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. 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 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.

NOTE

If a class is immutable and the cost of computing the hash code is significant, you might consider caching the hash code in the object rather than recalculating it each time it is requested.

// Lazily initialized, cached hashCode

private volatile int hashCode; // (See Item 71)

@Override public int hashCode() {

int result = hashCode;

if (result == 0) {

result = 17;

result = 31 * result + areaCode;

result = 31 * result + prefix;

result = 31 * result + lineNumber;

hashCode = result;

}

return result;

}

巴特西

Effective Java 09 Always override hashCode when you override equals

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