how to create an instance from a class that extends comparable and implement generic interface?

Question

I have a class that extends comparable and implements a generic interface and I want to create an instance of that class .

for example

interface MinMax<T extends Comparable<T>> {...}

class Employee<T extends Comparable<T>> implements MinMax<T> {...}

in this example i want to create an instance of Employee , how is that possible ?

new Employee<String>() or new Employer<Integer>, for example. You simply need to provide a type variable within the bounds. (Depending on the context, new Employee<>() might work too). — Andy Turner
– Andy Turner, Commented Sep 23, 2018 at 18:50

ernest_k · Accepted Answer · 2018-09-23 18:56:55Z

2

Employee doesn't need to take a type parameter. You want to make it Comparable and then use it as the type argument for MinMax:

class Employee implements Comparable<Employee>, MinMax<Employee> {
    @Override
    public int compareTo(Employee e) {
        //compare and return...
    }
}

With that, you don't need a type parameter for employee:

Employee e = new Employee(); //or whatever constructor you declared...

answered Sep 23, 2018 at 18:56

ernest_k

45.5k5 gold badges58 silver badges107 bronze badges

Sign up to request clarification or add additional context in comments.

Collectives™ on Stack Overflow

how to create an instance from a class that extends comparable and implement generic interface?

1 Answer 1

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Your Answer

Sign up or log in

Post as a guest

Related