Why can we not use default methods in lambda expressions?

Question

I was reading this tutorial on Java 8 where the writer showed the code:

interface Formula {
    double calculate(int a);

    default double sqrt(int a) {
           


        

Answer 1

Lambda expressions work in a completely different way from anonymous classes in that this represents the same thing that it would in the scope surrounding the expression.

For example, this compiles

class Main {

    public static void main(String[] args) {
        new Main().foo();
    }

    void foo() {
        System.out.println(this);
        Runnable r = () -> {
            System.out.println(this);
        };
        r.run();
    }
}

and it prints something like

Main@f6f4d33
Main@f6f4d33

In other words this is a Main, rather than the object created by the lambda expression.

So you cannot use sqrt in your lambda expression because the type of the this reference is not Formula, or a subtype, and it does not have a sqrt method.

Formula is a functional interface though, and the code

Formula f = a -> a;

compiles and runs for me without any problem.

Although you cannot use a lambda expression for this, you can do it using an anonymous class, like this:

Formula f = new Formula() { 
    @Override 
    public double calculate(int a) { 
        return sqrt(a * 100); 
    }
};