Given:
public class Foo {
@Override
public boolean equals(final Object obj) {
// implementation
}
@Override
public int hashCode() {
// implementation
}
}
and
public class Bar extends Foo {
@Override
public boolean equals(final Object obj) {
// different implementation
}
@Override
public int hashCode() {
// different implementation
}
}
I understand why Checkstyle gives me “Design for extension: the hashCode method” is not intended to be extended - it must be abstract, final or empty. ”The method is neither final, nor abstract, nor empty. But how else can I achieve this, but not violate any rules or guidelines of the TOE? For example, where will it be used:
Foo (for brevity, I used the default implementation of Object)
public class Foo {
public int getX() {
return x;
}
public void setX(final int x) {
this.x = x;
}
@Override
public int hashCode() {
return super.hashCode();
}
@Override
public boolean equals(final Object obj) {
return super.equals(obj);
}
private int x;
}
public class Bar extends Foo {
@Override
public int hashCode() {
final int prime = 31;
int result = 1;
result = result * prime + getX();
return result;
}
@Override
public boolean equals(final Object obj) {
if (obj == null) {
return false;
}
if (obj == this) {
return true;
}
if (!(obj instanceof Bar)) {
return false;
}
final Bar bar = (Bar) obj;
if (bar.getX() != this.getX()) {
return false;
}
return true;
}
}
public static void main(final String[] args) {
final Bar barOne = new Bar();
barOne.setX(1);
final Bar barTwo = new Bar();
barTwo.setX(1);
final Map<Bar, String> barMap = new HashMap<>();
barMap.put(barOne, null);
barMap.put(barTwo, null);
System.out.println(barMap.size());
final Foo fooOne = new Foo();
fooOne.setX(1);
final Foo fooTwo = new Foo();
fooTwo.setX(1);
final Map<Foo, String> fooMap = new HashMap<>();
fooMap.put(fooOne, null);
fooMap.put(fooTwo, null);
System.out.println(fooMap.size());
}
Outputs
1
2
, , Map , -. Object Object (Foo) (). Bar x. , , hashCode .
, , Foo Bar ?