如何重写equals（）方法

Question

public class Car
{
    private String name;  
    public int id;     


    public Car(String name, int id) 
    {
        this.name = name;
        this.id = id;
    }

@Override
public boolean equals(Object ob) 
{
    if (!(ob instanceof Car))
    {    
      return false;
    }
 Car that = (Car)ob;
 return this.id == that.id;
}

@Override
public int hashCode() 
{
    return id;
}
// this class also got getters and setters 

Then I got another class

public class CarList
{
        private Collection<Car> cars;

    public CarList()
    {
        cars = new HashSet<>();
    }


   public boolean insertCar(Car car)
    {
        return cars.add(car); 
    }

My question is: How to properly override equals() and hashCode() method, where I consider 'id' and 'name' attribute for object comparsion and hashCode calculation ( so there is no possibility to have 2 objects with the same name and ID - because in this code as it is - it only takes 'id' attribute for object comparsion)?

Answer 1

As of Java 7, there are static methods on Objects that makes implementing hashCode and equals easier. This should work well, assuming you don't want to use getClass() instead of instanceof to determine type compatibility. That depends on how subclasses of Car should compare to Car s.

@Override
public boolean equals(Object ob) 
{
    if (!(ob instanceof Car))
    {    
      return false;
    }
 Car that = (Car)ob;
 return Objects.equals(this.id, that.id) && Objects.equals(this.name, that.name);
}

@Override
public int hashCode() 
{
    return Objects.hash(id, name);
}

Answer 2

Instead of using

if (!(ob instanceof Car))
{    
  return false;
}

You should think about using

if (getClass() != obj.getClass())
  return false;

Lets assume you have ForWdCar extends Car and TwoWdCar extends Car with equal name and id.

Do you want them to be equal? 1st solution,

Do you want them to be unequal? 2nd solution

You don't care, such cases don't happen? Second solution, it's faster.