How to make an object of association comparable by only one generic parameter type

Question

I'm trying to understand what <K, V extends Comparable<? super V>> actually means.

If compareTo() is called to compare two Association<K,V> objects; which of the two generic types K or V will be used to make the comparison?

What will happen when <K, V extends Comparable<? super K>> is used instead?

public class Association<K, V extends Comparable<? super V>>
{
	K key;
	V value;
	//---------------------------------------
  
        public int compareTo(Association<K, V> object) 
	{
		return this.key.compareTo(object.key);
	}
}

edited the compareTo() method.

Answer 1

First to explain your current definition:

public class Association<K, V extends Comparable<? super V>>

This means you are defining a public class called Association This class has 2 type parameters: K and V. Of these, V has the additional restraint that it must be comparable to ? super V (= V or a super type of V).

Since you want ot compare by the key, it should be K that is Comparable. Additionally, you want to compare the Associations themselves, so it too must be Comparable.

So change your class definition to the following:

public class Association<K extends Comparable<K>, V> implements Comparable<Association<K, V>>

Your compareTo would then be:

public int compareTo(Association<K, V> other) {
    return key.compareTo(other.key);
}