Order guarantees using streams and reducing chain of consumers

Question

So as it goes in the current scenario, we have a set of APIs as listed below:

Consumer start();
Consumer performDailyAggregates();
Consumer         


        

Answer 1

As Andreas pointed out, Consumer::andThen is an associative function and while the resulting consumer may have a different internal structure, it is still equivalent.

But let's debug it

public static void main(String[] args) {
    performAllTasks(IntStream.range(0, 10)
        .mapToObj(i -> new DebuggableConsumer(""+i)), new Object());
}
private static <T> void performAllTasks(Stream<Consumer<T>> consumerList, T data) {
    Consumer<T> reduced = consumerList.reduce(Consumer::andThen).orElse(x -> {});
    reduced.accept(data);
    System.out.println(reduced);
}
static class DebuggableConsumer implements Consumer<Object> {
    private final Consumer<Object> first, second;
    private final boolean leaf;
    DebuggableConsumer(String name) {
        this(x -> System.out.println(name), x -> {}, true);
    }
    DebuggableConsumer(Consumer<Object> a, Consumer<Object> b, boolean l) {
        first = a; second = b;
        leaf = l;
    }
    public void accept(Object t) {
        first.accept(t);
        second.accept(t);
    }
    @Override public Consumer<Object> andThen(Consumer<? super Object> after) {
        return new DebuggableConsumer(this, after, false);
    }
    public @Override String toString() {
        if(leaf) return first.toString();
        return toString(new StringBuilder(200), 0, 0).toString();
    }
    private StringBuilder toString(StringBuilder sb, int preS, int preEnd) {
        int myHandle = sb.length()-2;
        sb.append(leaf? first: "combined").append('\n');
        if(!leaf) {
            int nPreS=sb.length();
            ((DebuggableConsumer)first).toString(
                sb.append(sb, preS, preEnd).append("\u2502 "), nPreS, sb.length());
            nPreS=sb.length();
            sb.append(sb, preS, preEnd);
            int lastItemHandle=sb.length();
            ((DebuggableConsumer)second).toString(sb.append("  "), nPreS, sb.length());
            sb.setCharAt(lastItemHandle, '\u2514');
        }
        if(myHandle>0) {
            sb.setCharAt(myHandle, '\u251c');
            sb.setCharAt(myHandle+1, '\u2500');
        }
        return sb;
    }
}

will print

0
1
2
3
4
5
6
7
8
9
combined
├─combined
│ ├─combined
│ │ ├─combined
│ │ │ ├─combined
│ │ │ │ ├─combined
│ │ │ │ │ ├─combined
│ │ │ │ │ │ ├─combined
│ │ │ │ │ │ │ ├─combined
│ │ │ │ │ │ │ │ ├─SO$DebuggableConsumer$$Lambda$21/0x0000000840069040@378fd1ac
│ │ │ │ │ │ │ │ └─SO$DebuggableConsumer$$Lambda$21/0x0000000840069040@49097b5d
│ │ │ │ │ │ │ └─SO$DebuggableConsumer$$Lambda$21/0x0000000840069040@6e2c634b
│ │ │ │ │ │ └─SO$DebuggableConsumer$$Lambda$21/0x0000000840069040@37a71e93
│ │ │ │ │ └─SO$DebuggableConsumer$$Lambda$21/0x0000000840069040@7e6cbb7a
│ │ │ │ └─SO$DebuggableConsumer$$Lambda$21/0x0000000840069040@7c3df479
│ │ │ └─SO$DebuggableConsumer$$Lambda$21/0x0000000840069040@7106e68e
│ │ └─SO$DebuggableConsumer$$Lambda$21/0x0000000840069040@7eda2dbb
│ └─SO$DebuggableConsumer$$Lambda$21/0x0000000840069040@6576fe71
└─SO$DebuggableConsumer$$Lambda$21/0x0000000840069040@76fb509a

whereas changing the reduction code to

private static <T> void performAllTasks(Stream<Consumer<T>> consumerList, T data) {
    Consumer<T> reduced = consumerList.parallel().reduce(Consumer::andThen).orElse(x -> {});
    reduced.accept(data);
    System.out.println(reduced);
}

prints on my machine

0
1
2
3
4
5
6
7
8
9
combined
├─combined
│ ├─combined
│ │ ├─SO$DebuggableConsumer$$Lambda$22/0x0000000840077c40@49097b5d
│ │ └─SO$DebuggableConsumer$$Lambda$22/0x0000000840077c40@6e2c634b
│ └─combined
│   ├─SO$DebuggableConsumer$$Lambda$22/0x0000000840077c40@37a71e93
│   └─combined
│     ├─SO$DebuggableConsumer$$Lambda$22/0x0000000840077c40@7e6cbb7a
│     └─SO$DebuggableConsumer$$Lambda$22/0x0000000840077c40@7c3df479
└─combined
  ├─combined
  │ ├─SO$DebuggableConsumer$$Lambda$22/0x0000000840077c40@7106e68e
  │ └─SO$DebuggableConsumer$$Lambda$22/0x0000000840077c40@7eda2dbb
  └─combined
    ├─SO$DebuggableConsumer$$Lambda$22/0x0000000840077c40@6576fe71
    └─combined
      ├─SO$DebuggableConsumer$$Lambda$22/0x0000000840077c40@76fb509a
      └─SO$DebuggableConsumer$$Lambda$22/0x0000000840077c40@300ffa5d

illustrating the point of Andreas’ answer, but also highlighting an entirely different problem. You may max it out by using, e.g. IntStream.range(0, 100) in the example code.

The result of the parallel evaluation is actually better than the sequential evaluation, as the sequential evaluation creates an unbalanced tree. When accepting an arbitrary stream of consumers, this can be an actual performance issue or even lead to a StackOverflowError when trying to evaluate the resulting consumer.

For any nontrivial number of consumers, you actually want a balanced consumer tree, but using a parallel stream for that is not the right solution, as a) Consumer::andThen is a cheap operation with no real benefit from parallel evaluation and b) the balancing would depend on unrelated properties, like the nature of the stream source and the number of CPU cores, which determine when the reduction falls back to the sequential algorithm.

Of course, the simplest solution would be

private static <T> void performAllTasks(Stream<Consumer<T>> consumers, T data) {
    consumers.forEachOrdered(c -> c.accept(data));
}

But when you want to construct a compound Consumer for re-using, you may use

private static final int ITERATION_THRESHOLD = 16; // tune yourself

public static <T> Consumer<T> combineAllTasks(Stream<Consumer<T>> consumers) {
    List<Consumer<T>> consumerList = consumers.collect(Collectors.toList());
    if(consumerList.isEmpty()) return t -> {};
    if(consumerList.size() == 1) return consumerList.get(0);
    if(consumerList.size() < ITERATION_THRESHOLD)
        return balancedReduce(consumerList, Consumer::andThen, 0, consumerList.size());
    return t -> consumerList.forEach(c -> c.accept(t));
}
private static <T> T balancedReduce(List<T> l, BinaryOperator<T> f, int start, int end) {
    if(end-start>2) {
        int mid=(start+end)>>>1;
        return f.apply(balancedReduce(l, f, start, mid), balancedReduce(l, f, mid, end));
    }
    T t = l.get(start++);
    if(start<end) t = f.apply(t, l.get(start));
    assert start==end || start+1==end;
    return t;
}

The code will provide a single Consumer just using a loop when the number of consumers exceeds a threshold. This is the simplest and most efficient solution for a larger number of consumers and in fact, you could drop all other approaches for the smaller numbers and still get a reasonable performance…

Note that this still doesn’t hinder parallel processing of the stream of consumers, if their construction really benefits from it.

Answer 2

Even if the Stream<Consumer<T>> is made parallel, the resulting compound Consumer will execute the individual consumers in order, assuming:

• The Stream is ordered.
  A stream sourced by a List is ordered, even with parallel enabled.

• The accumulator passed to reduce() is associative.
  Consumer::andThen is associative.

Let's say you have a list of 4 consumers [A, B, C, D]. Normally, without parallel, the following would happen:

x = A.andThen(B);
x = x.andThen(C);
compound = x.andThen(D);

so that calling compound.apply() would call A, B, C, then D in that order.

If you enable parallel, the stream framework might instead split that to be processed by 2 threads, [A, B] by thread 1, and [C, D] by thread 2.

That means the following will happen:

x = A.andThen(B);
y = C.andThen(D);
compound = x.andThen(y);

The result is that x is applied first, which means A then B, then y is applied, which means C then D.

So although the compound consumer is built like [[A, B], [C, D]] instead of the left-associative [[[A, B], C], D], the 4 consumers are executed in order, all because Consumer::andThen is associative.