如何通过单个参数传递Java 8 lambda [英] How to pass a Java 8 lambda with a single parameter

问题描述

我只想传递一个lambda(代码块)并在需要时执行它.如何在下面的代码中实现方法executeLambda(...)(以及什么是方法签名):

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();
}

推荐答案

您的lambda带有一个参数，但是您只能将lambda传递给executeLambda，而不是值.如果您想让lambda捕获局部变量，请不要将其写为带参数，但是如果您确实希望它使用一个参数，则可以这样写:

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);
}

如果希望它捕获value，请使用Runnable，将lambda写为() -> print(value)，然后像runnable.run()那样调用它.

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

这篇关于如何通过单个参数传递Java 8 lambda的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！