Java Pass Method as Parameter

Question

I am looking for a way to pass a method by reference. I understand that Java does not pass methods as parameters, however, I would like to get an alternative.

I\'ve

Answer 1

In Java 8, you can now pass a method more easily using Lambda Expressions and Method References. 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);

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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.

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

interface I {
    public void myMethod(Component component);
}

class A {
    public void changeColor(Component component) {
        // code here
    }

    public void changeSize(Component component) {
        // code here
    }
}

class B {
    public void setAllComponents(Component[] myComponentArray, I myMethodsInterface) {
        for(Component leaf : myComponentArray) {
            if(leaf instanceof Container) { // recursive call if Container
                Container node = (Container)leaf;
                setAllComponents(node.getComponents(), myMethodInterface);
            } // end if node
            myMethodsInterface.myMethod(leaf);
        } // end looping through components
    }
}

class C {
    A a = new A();
    B b = new B();

    public C() {
        b.setAllComponents(this.getComponents(), component -> a.changeColor(component));
        b.setAllComponents(this.getComponents(), component -> a.changeSize(component));
    }
}

Class C can be shortened even a bit further by the use of method references like so:

class C {
    A a = new A();
    B b = new B();

    public C() {
        b.setAllComponents(this.getComponents(), a::changeColor);
        b.setAllComponents(this.getComponents(), a::changeSize);
    }
}