Question

I have this statement about java equals and hashcode.

if we were to use such an Integer object for a key in a HashMap, we would not be able to reliably retrieve the associated value

what it means ? why it says ‘not reliably’ ? I write a test program, it always works.

public class Test1 {

    public static void main(String[] args){
        Map<Integer, Student> map = new HashMap<Integer, Student>();

        map.put(1, new Student("john"));        
        map.put(2, new Student("peter"));

        Student s1 = map.get(1);
        Student s2 = map.get(1);
        Student s3 = map.get(2);
        System.out.println("s1:"+s1+" s2:"+s2+" s3:"+s3);

        System.out.println(s1==s2);
        System.out.println(s1==s3);
        System.out.println(s1.equals(s3));
        System.out.println(s1.equals(s2));
    }   
}

class Student{
    private String name;
    public Student(String name){        
        this.name = name;       
    }

    public String getName(){
        return this.name;
    }   
}

Answer 1

You have misunderstood! Read the whole paragraph:

What would happen if Integer did not override equals() and hashCode()?
Nothing, if we never used an Integer as a key in a HashMap or other
hash-based collection. However, if we were to use such an Integer
object for a key in a HashMap, we would not be able to reliably
retrieve the associated value, unless we used the exact same Integer
instance in the get() call as we did in the put() call.

http://www.ibm.com/developerworks/java/library/j-jtp05273/index.html

This is saying that the Integer class must have a hashCode() method and equals() in order to work. It does have those methods, so it’s fine. It will work.

The example was saying that references to objects that store values should not be compared because two Integers with the same integer value may actually be different objects.

Integer x = new Integer(5);
Integer y = new Integer(5);

x.equals(y) is always true. x == y is not always true (but can be sometimes). The rules change for different ranges of value. Never rely on == for any object references unless you are completely sure you know what you are doing.

Also note that you are passing int arguments (primitive types) not Integers (references to objects) but your collection’s generic type is Integer. This means there is also boxing to throw into the equation!