How to pass a Java 8 lambda with a single parameter

Question

I want to simply pass a lambda (chunk of code) and execute it when I need to. How do I implement the method executeLambda(...) in the code below (as well what is the method signature):

public static void main(String[] args)
{
    String value = "Hello World";
    executeLambda(value -> print(value));
}

public static void print(String value)
{
    System.out.println(value);
}

public static void executeLambda(lambda)
{
    someCode();
    lamda.executeLambdaCode();
    someMoreCode();
}

Answer 1

Your lambda takes one parameter, but you only pass the lambda to executeLambda, not the value. If you want the lambda to capture the local variable, don't write it taking a parameter, but if you do really want it to take one parameter, you would write it like this:

import java.util.function.Consumer;

public static void main(String[] args) {
    String message = "Hello World";
    executeLambda(message, value -> print(value));
}

public static void executeLambda(String value, Consumer<String> lambda) {
    lambda.accept(value);
}

If you want it to capture the value, then use Runnable, write the lambda as () -> print(value), and call it like runnable.run().

Answer 2

public static void main(String[] args)
{
    String value = "Hello World";
    executeLambda(() -> print(value));
}

public static void print(String value)
{
    System.out.println(value);
}

public static void executeLambda(Runnable runnable)
{
    runnable.run();
}