How do I define a method which takes a lambda as a parameter in Java 8?

Question

In Java 8, methods can be created as Lambda expressions and can be passed by reference (with a little work under the hood). There are plenty of examples online with lambdas being created and used with methods, but no examples of how to make a method taking a lambda as a parameter. What is the syntax for that?

MyClass.method((a, b) -> a + b);
    

class MyClass{
  //How do I define this method?
  static int method(Lambda l) {
    return l(5, 10);
  }
}

Answer 1

Lambdas are purely a call-site construct: the recipient of the lambda does not need to know that a Lambda is involved, instead it accepts an Interface with the appropriate method.

In other words, you define or use a functional interface (i.e. an interface with a single method) that accepts and returns exactly what you want.

Since Java 8 there is a set of commonly-used interface types in java.util.function.

For this specific use case there's java.util.function.IntBinaryOperator with a single int applyAsInt(int left, int right) method, so you could write your method like this:

static int method(IntBinaryOperator op){
    return op.applyAsInt(5, 10);
}

But you can just as well define your own interface and use it like this:

public interface TwoArgIntOperator {
    public int op(int a, int b);
}

//elsewhere:
static int method(TwoArgIntOperator operator) {
    return operator.op(5, 10);
}

Then call the method with a lambda as parameter:

public static void main(String[] args) {
    TwoArgIntOperator addTwoInts = (a, b) -> a + b;
    int result = method(addTwoInts);
    System.out.println("Result: " + result);
}

Using your own interface has the advantage that you can have names that more clearly indicate the intent.