[java] How to pass a function as a parameter in Java?

Lambda Expressions

To add on to jk.'s excellent answer, you can now pass a method more easily using Lambda Expressions (in Java 8). First, some background. A functional interface is an interface that has one and only one abstract method, although it can contain any number of default methods (new in Java 8) and static methods. A lambda expression can quickly implement the abstract method, without all the unnecessary syntax needed if you don't use a lambda expression.

Without lambda expressions:

obj.aMethod(new AFunctionalInterface() {
    @Override
    public boolean anotherMethod(int i)
    {
        return i == 982
    }
});

With lambda expressions:

obj.aMethod(i -> i == 982);

Here is an excerpt from the Java tutorial on Lambda Expressions:

Syntax of Lambda Expressions

A lambda expression consists of the following:

  • A comma-separated list of formal parameters enclosed in parentheses. The CheckPerson.test method contains one parameter, p, which represents an instance of the Person class.

    Note: You can omit the data type of the parameters in a lambda expression. In addition, you can omit the parentheses if there is only one parameter. For example, the following lambda expression is also valid:

    p -> p.getGender() == Person.Sex.MALE 
        && p.getAge() >= 18
        && p.getAge() <= 25
    
  • The arrow token, ->

  • A body, which consists of a single expression or a statement block. This example uses the following expression:

    p.getGender() == Person.Sex.MALE 
        && p.getAge() >= 18
        && p.getAge() <= 25
    

    If you specify a single expression, then the Java runtime evaluates the expression and then returns its value. Alternatively, you can use a return statement:

    p -> {
        return p.getGender() == Person.Sex.MALE
            && p.getAge() >= 18
            && p.getAge() <= 25;
    }
    

    A return statement is not an expression; in a lambda expression, you must enclose statements in braces ({}). However, you do not have to enclose a void method invocation in braces. For example, the following is a valid lambda expression:

    email -> System.out.println(email)
    

Note that a lambda expression looks a lot like a method declaration; you can consider lambda expressions as anonymous methods—methods without a name.


Here is how you can "pass a method" using a lambda expression:

Note: this uses a new standard functional interface, java.util.function.IntConsumer.

class A {
    public static void methodToPass(int i) { 
        // do stuff
    }
}
import java.util.function.IntConsumer;

class B {
    public void dansMethod(int i, IntConsumer aMethod) {
        /* you can now call the passed method by saying aMethod.accept(i), and it
        will be the equivalent of saying A.methodToPass(i) */
    }
}
class C {
    B b = new B();

    public C() {
        b.dansMethod(100, j -> A.methodToPass(j));   //Lambda Expression here
    }
}

The above example can be shortened even more using the :: operator.

public C() {
    b.dansMethod(100, A::methodToPass);
}