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.

For this Java 8 comes with a set of commonly-used interface types in java.util.function (thanks to Maurice Naftalin for the hint about the JavaDoc).

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.