Instance of runnable from a member function

Question

class test
{
        public static void main(String[] args)
        {
                System.out.println("Start");
                MyClass myClass = new MyClass();
                myClass.runFunc();
                System.out.println("End");
        }
}

class MyClass
{
        public void hello()
        {
                System.out.println("Hello");
        }
        public void runFunc()
        {
                Runnable run = this::hello;
                new Thread(run).start();
        }
}

Here hello is a member function of MyClass not an instance of any class which implements Runnble. But still the assignment Runnable run = this::hello; doesn't fail and the hello functions is getting executed in the context of the a new thread.

I know a Runnable can can be also initialized with a lambda expression but it's not the case here either.

I am new to Java. Can somebody please explain how this is working and what is the logic behind this behavior.

Answer 1

This line:

Runnable run = this::hello;

is lambda short-hand, which is really:

Runnable run = new Runnable() {
    @Override
    public void run() {
        MyClass.this.hello();
    }
}

You're not creating a Runnable "of" your member function, but rather constructing a new Runnable, with a run() method that calls your member function.

The following would also be functionally equivalent:

Runnable run = () -> this.hello();