Lambda表达式和高阶函数 [英] Lambda expressions and higher-order functions

问题描述

如何使用Java 8编写Java 8，使用闭包支持一个方法，它接受
作为参数，并返回一个函数作为值？

How can I write with Java 8 with closures support a method that take as argument a function and return function as value?

推荐答案

在Java Lambda API中，主类是 java.util.function.Function 。

In Java Lambda API the main class is java.util.function.Function.

您可以使用此接口的引用

You can use a reference to this interface in the same way as you would do with all other references: create that as variable, return it as a result of computation and so on.

这是一个非常简单的例子，它可能会帮助你you：

Here is quite simple example which might help you:

    public class HigherOrder {

        public static void main(String[] args) {
            Function<Integer, Long> addOne = add(1L);

            System.out.println(addOne.apply(1)); //prints 2

            Arrays.asList("test", "new")
                    .parallelStream()  // suggestion for execution strategy
                    .map(camelize)     // call for static reference
                    .forEach(System.out::println);
        }

        private static Function<Integer, Long> add(long l) {
            return (Integer i) -> l + i;
        }

        private static Function<String, String> camelize = (str) -> str.substring(0, 1).toUpperCase() + str.substring(1);
    }

如果您需要传递多于1个参数，请查看 compose 方法，但它的使用是很棘手的。

If you need to pass more then 1 parameter, please take a look into compose method, but its usage is quite tricky.

一般来说，从Java的闭包和lambdas基本上是语法-sugar，他们似乎没有所有的功能编程能力。

In general from my opinion closures and lambdas in Java is basically syntax-sugar, and they seem to not have all capabilities of functional programming.

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