如果我在并行使用lambda则会发生死锁，但是如果我使用匿名类却不会发生死锁？

Question

以下代码导致死锁（在我的电脑上）：

public class Test {
    static {
        final int SUM = IntStream.range(0, 100)
                .parallel()
                .reduce((n, m) -> n + m)
                .getAsInt();
    }

    public static void main(String[] args) {
        System.out.println("Finished");
    }
}

但是，如果我将reducelambda参数替换为匿名类，则不会导致死锁：

public class Test {
    static {
        final int SUM = IntStream.range(0, 100)
                .parallel()
                .reduce(new IntBinaryOperator() {
                    @Override
                    public int applyAsInt(int n, int m) {
                        return n + m;
                    }
                })
                .getAsInt();
    }

    public static void main(String[] args) {
        System.out.println("Finished");
    }
}

你能解释一下这种情况吗？

P.S.

我发现该代码（与之前的代码有些不同）：

public class Test {
    static {
        final int SUM = IntStream.range(0, 100)
                .parallel()
                .reduce(new IntBinaryOperator() {
                    @Override
                    public int applyAsInt(int n, int m) {
                        return sum(n, m);
                    }
                })
                .getAsInt();
    }

    private static int sum(int n, int m) {
        return n + m;
    }

    public static void main(String[] args) {
        System.out.println("Finished");
    }
}

工作不稳定。在大多数情况下，它挂起了，但是有时它成功完成了：

在此处输入图片说明

我真的不明白为什么这种行为不稳定。实际上，我重新测试了第一个代码段，并且行为相同。因此，最新的代码等于第一个。

为了了解使用了哪些线程，我在“日志记录”后面添加了以下内容：

public class Test {
    static {
        final int SUM = IntStream.range(0, 100)
                .parallel()
                .reduce((n, m) -> {
                    System.out.println(Thread.currentThread().getName());
                    return (n + m);
                })
                .getAsInt();
    }

    public static void main(String[] args) {
        System.out.println("Finished");
    }
}

对于应用程序成功完成的情况，我会看到以下日志：

main
main
main
main
ForkJoinPool.commonPool-worker-1
ForkJoinPool.commonPool-worker-1
ForkJoinPool.commonPool-worker-1
ForkJoinPool.commonPool-worker-1
ForkJoinPool.commonPool-worker-1
ForkJoinPool.commonPool-worker-1
ForkJoinPool.commonPool-worker-1
ForkJoinPool.commonPool-worker-1
ForkJoinPool.commonPool-worker-1
ForkJoinPool.commonPool-worker-1
ForkJoinPool.commonPool-worker-1
ForkJoinPool.commonPool-worker-1
ForkJoinPool.commonPool-worker-1
ForkJoinPool.commonPool-worker-1
ForkJoinPool.commonPool-worker-1
ForkJoinPool.commonPool-worker-1
ForkJoinPool.commonPool-worker-1
ForkJoinPool.commonPool-worker-1
ForkJoinPool.commonPool-worker-1
ForkJoinPool.commonPool-worker-1
ForkJoinPool.commonPool-worker-1
ForkJoinPool.commonPool-worker-1
ForkJoinPool.commonPool-worker-1
ForkJoinPool.commonPool-worker-1
main
ForkJoinPool.commonPool-worker-1
ForkJoinPool.commonPool-worker-1
ForkJoinPool.commonPool-worker-1
ForkJoinPool.commonPool-worker-1
ForkJoinPool.commonPool-worker-1
ForkJoinPool.commonPool-worker-1
ForkJoinPool.commonPool-worker-1
ForkJoinPool.commonPool-worker-1
ForkJoinPool.commonPool-worker-1
ForkJoinPool.commonPool-worker-1
ForkJoinPool.commonPool-worker-1
ForkJoinPool.commonPool-worker-1
main
main
main
main
main
main
main
main
main
main
main
main
main
main
main
main
main
ForkJoinPool.commonPool-worker-1
ForkJoinPool.commonPool-worker-1
ForkJoinPool.commonPool-worker-1
ForkJoinPool.commonPool-worker-1
ForkJoinPool.commonPool-worker-1
main
main
main
main
main
main
main
main
main
main
main
main
main
main
main
main
main
main
main
main
main
main
main
main
main
main
main
ForkJoinPool.commonPool-worker-1
ForkJoinPool.commonPool-worker-1
ForkJoinPool.commonPool-worker-1
ForkJoinPool.commonPool-worker-1
ForkJoinPool.commonPool-worker-1
ForkJoinPool.commonPool-worker-1
ForkJoinPool.commonPool-worker-1
ForkJoinPool.commonPool-worker-1
ForkJoinPool.commonPool-worker-1
Finished
PS 2

我不明白reduce是足够复杂的操作。我找到了一个更简单的示例来显示该问题：

public class Test {
    static {
        System.out.println("static initializer: " + Thread.currentThread().getName());

        final long SUM = IntStream.range(0, 2)
                .parallel()
                .mapToObj(i -> {
                    System.out.println("map: " + Thread.currentThread().getName() + " " + i);
                    return i;
                })
                .count();
    }

    public static void main(String[] args) {
        System.out.println("Finished");
    }
}

对于幸福的情况（罕见的情况），我看到以下输出：

static initializer: main
map: main 1
map: main 0
Finished

扩展流范围的幸福案例示例：

static initializer: main
map: main 2
map: main 3
map: ForkJoinPool.commonPool-worker-2 4
map: ForkJoinPool.commonPool-worker-1 1
map: ForkJoinPool.commonPool-worker-3 0
Finished

导致死锁的案例示例：

static initializer: main
map: main 1

它还会导致死锁，但不会导致每次启动。

Answer 1

区别在于lambda主体是在同一Test类中编写的，即合成方法

private static int lambda$static$0(int n, int m) {
    return n + m;
}

在第二种情况下，接口的实现位于不同的 Test$1类中。因此，并行流的线程不会调用的静态方法，Test因此不依赖于Test初始化。