Implicit type conversion for lambda expression

Question

Consider the following piece of class:

public void method() {
    test(() -> { });
}

void test(Runnable a) {
    System.out.println("Test 1");
}

void test(A a) {
    System.out.println("Test 2");
}

interface A extends Runnable {

}

Invoking method method() will lead to Test 2 output. This means, that lambda expression () -> { } was implicitly converted to A . Why?

Answer 1

It's the same standard rule applied to all overloads. Java will choose the most specific applicable method.

Both methods accept an argument that is of a functional interface type. The lambda expression

() -> { }

is convertible to both those types. A is a subclass of Runnable and is therefore more specific. The method with a parameter type of A therefore gets chosen.