该方法引用的有效方法是什么？幕后发生了什么？

Question

我有一个以下程序

import java.util.Scanner;

@FunctionalInterface
interface Retry {

    int run(Runnable action, int maxAttempts, long delayBeforeRetryMs);
}

final class RetryUtils {
    public static Retry retry= RetryUtils::retryAction; // assign the retryAction method to this variable

    private RetryUtils() { }

    public static int retryAction(
            Runnable action, int maxAttempts, long delayBeforeRetryMs) {

        int fails = 0;
        while (fails < maxAttempts) {
            try {
                action.run();
                return fails;
            } catch (Exception e) {
                System.out.println("Something goes wrong");
                fails++;
                try {
                    Thread.sleep(delayBeforeRetryMs);
                } catch (InterruptedException interruptedException) {
                    throw new RuntimeException(interruptedException);
                }
            }
        }
        return fails;
    }
}

class Retrying {
    private static final int MAX_ATTEMPTS = 10;
    private static final long DELAY_MS = 1;

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        RetryUtils.retry.run(() -> System.out.println(scanner.nextLine()), MAX_ATTEMPTS, DELAY_MS);
    }
}

我看到方法retryAction返回int。那么公共静态Retry retry=RetryUtils::retryAction；是如何对Retry类型的对象进行有效分配的？这是如何编译的？幕后发生了什么？

Answer 1

期望int作为返回类型不是函数调用。它是一个lambda表达式，它创建了Retry接口的匿名实现。它几乎等于这个

 public static Retry retry= new Retry(){
     int run(Runnable action, int maxAttempts, long delayBeforeRetryMs)
     RetryUtils.this.retryAction(action,maxAttemps,delayBeforeRetryMs)
}

由于静力学的原因，这个类比几乎是准确的，但你应该明白这个道理。