Question

我想走一个异步函数的搜索空间。我将逻辑编码如下：

/**
 * Assuming that a function maps a range of inputs to the same output value, minimizes the input value while
 * maintaining the output value.
 *
 * @param previousInput the last input known to return {@code target}
 * @param currentInput  the new input value to evaluate
 * @param function      maps an input to an output value
 * @param target        the expected output value
 * @return the minimum input value that results in the {@code target} output value
 * <br>{@code @throws NullPointerException} if any argument is null
 * <br>{@code @throws IllegalArgumentException} if {@code stepSize} is zero}
 */
private static CompletionStage<BigDecimal> optimizeInput(BigDecimal previousInput,
                                                         BigDecimal currentInput,
                                                         BigDecimal stepSize,
                                                         Function<BigDecimal, CompletionStage<BigDecimal>> function,
                                                         BigDecimal target)
{
    return function.apply(currentInput).thenCompose(output ->
    {
        assertThat("stepSize", stepSize).isNotZero();
        int outputMinusTarget = output.compareTo(target);
        if (outputMinusTarget != 0)
            return CompletableFuture.completedFuture(previousInput);
        BigDecimal nextInput = currentInput.add(stepSize);
        if (nextInput.compareTo(BigDecimal.ZERO) < 0)
            return CompletableFuture.completedFuture(previousInput);
        return optimizeInput(currentInput, nextInput, stepSize, function, target);
    });
}

不幸的是，如果函数有一个很大的搜索空间，这会在一些迭代后引发StackoverflowError。是否可以使用固定大小的堆栈迭代地遍历搜索空间？

Answer 1

你有以下递归结构

CompletableFuture<T> compute(...) {
  return asyncTask().thenCompose(t -> {
    if (...)
      return completedFuture(t);
    } else {
      return compute(...);
    }
  }
}

您可以重写它，避免在完成期间完成未来的组合及其堆栈使用。

CompletableFuture<T> compute(...) {
  CompletableFuture<T> result = new CompletableFuture<>();
  computeHelper(result, ...);
  return result;
}   

void computeHelper(CompletableFuture<T> result, ...) {
  asyncTask().thenAccept(t -> {
    if (...) {
      result.complete(t);
    } else {
      computeHelper(result, ...);
    }
  });
}

如果asyncTask()实际上不是异步并只使用当前线程，则必须将thenAccept替换为其异步版本之一，以使用执行程序任务队列而不是线程堆栈。

Answer 2

dfogni的答案应该可以正常工作 - 但是为了完整性，在方法使用trampolining类型技术同步的情况下，可以避免执行执行程序切换。

为了更容易，我引入了一个类，它捕获在迭代之间发生变化的状态，并引入实现完成检查并生成下一个状态的方法。我相信这与你原来的逻辑相同，但你可以三重检查。

private static CompletionStage<BigDecimal> optimizeInput(BigDecimal previousInput,
                                                          BigDecimal currentInput,
                                                          BigDecimal stepSize,
                                                          Function<BigDecimal, CompletionStage<BigDecimal>> function,
                                                          BigDecimal target) {
    class State {
        BigDecimal prev;
        BigDecimal curr;
        BigDecimal output;

        State(BigDecimal prev, BigDecimal curr, BigDecimal output) {
            this.prev = prev;
            this.curr = curr;
            this.output = output;
        }

        boolean shouldContinue() {
            return output.compareTo(target) == 0 && curr.add(stepSize).compareTo(BigDecimal.ZERO) >= 0;
        }

        CompletionStage<State> next() {
            BigDecimal nextInput = curr.add(stepSize);
            return function.apply(nextInput).thenApply(nextOutput -> new State(curr, nextInput, nextOutput));
        }
    }

    /* Now it gets complicated... we have to check if we're running on the same thread we were called on. If we
     * were, instead of recursively calling `next()`, we'll use PassBack to pass our new state back 
     * to the stack that called us.
     */
    class Passback {
        State state = null;
        boolean isRunning = true;

        State poll() {
            final State c = this.state;
            this.state = null;
            return c;
        }
    }
    class InputOptimizer extends CompletableFuture<BigDecimal> {
        void optimize(State state, final Thread previousThread, final Passback previousPassback) {
            final Thread currentThread = Thread.currentThread();

            if (currentThread.equals(previousThread) && previousPassback.isRunning) {
                // this is a recursive call, our caller will run it
                previousPassback.state = state;
            } else {
                Passback passback = new Passback();
                State curr = state;
                do {
                    if (curr.shouldContinue()) {
                        curr.next().thenAccept(next -> optimize(next, currentThread, passback));
                    } else {
                        complete(curr.prev);
                        return;
                    }
                // loop as long as we're making synchronous recursive calls
                } while ((curr = passback.poll()) != null);
                passback.isRunning = false;
            }
        }
    }

    InputOptimizer ret = new InputOptimizer();
    function.apply(currentInput)
            .thenAccept(output -> ret.optimize(
                    new State(previousInput, currentInput, output),
                    null, null));
    return ret;
}

好的，所以这很复杂。另外，请注意，这要求您的函数永远不会抛出异常或异常完成，这可能会有问题。您可以对此进行生成，因此您只需编写一次（正确的异常处理），可以在asyncutil library中找到（免责声明：我是此库的共同作者））。可能还有其他具有类似功能的库，很可能是像Rx这样的成熟反应库。使用asyncutil，

 private static CompletionStage<BigDecimal> optimizeInput(BigDecimal previousInput,
                                                          BigDecimal currentInput,
                                                          BigDecimal stepSize,
                                                          Function<BigDecimal, CompletionStage<BigDecimal>> function,
                                                          BigDecimal target) {
    // ... State class from before
    return function
            .apply(currentInput)
            .thenCompose(output -> AsyncTrampoline.asyncWhile(
                    State::shouldContinue, 
                    State::next, 
                    new State(previousInput, currentInput, output)))
            .thenApply(state -> state.prev);    
}

如何在不冒StackOverflowError风险的情况下使用CompletableFuture？

2 个答案: