Question

我需要为Web服务器实现全局对象收集统计信息。我有Statistics单身人士，其方法为addSample(long sample)，随后会调用updateMax。这显然是线程安全的。我有这种方法来更新整个统计信息的最大值：

AtomicLong max;

private void updateMax(long sample) {
    while (true) {
        long curMax = max.get();
        if (curMax < sample) {
            boolean result = max.compareAndSet(curMax, sample);
            if (result) break;
        } else {
            break;
        }
    }
}

此实施是否正确？我正在使用java.util.concurrent，因为我相信它会比简单的synchronized更快。是否有其他/更好的方法来实现这一点？

Answer 1

我认为这是正确的，但为了清晰起见，我可能会重写一点，肯定会添加评论：

private void updateMax(long sample) {
    while (true) {
        long curMax = max.get();
        if (curMax >= sample) {
            // Current max is higher, so whatever other threads are
            // doing, our current sample can't change max.
            break;
        }

        // Try updating the max value, but only if it's equal to the
        // one we've just seen. We don't want to overwrite a potentially
        // higher value which has been set since our "get" call.
        boolean setSuccessful = max.compareAndSet(curMax, sample);

        if (setSuccessful) {
            // We managed to update the max value; no other threads
            // got in there first. We're definitely done.
            break;
        }

        // Another thread updated the max value between our get and
        // compareAndSet calls. Our sample can still be higher than the
        // new value though - go round and try again.
    }
}

编辑：通常我至少首先尝试同步版本，只有在我发现它导致问题时才会使用这种无锁代码。

Answer 2

自Java 8起，引入了LongAccumulator。建议为

当多个线程时，此类通常优于AtomicLong 更新用于收集等目的的公共值统计，而不是细粒度的同步控制。低下更新争用，这两个类具有相似的特征。但在高争用的情况下，这个类的预期吞吐量是以更高的空间消耗为代价，显着提高。

您可以按如下方式使用它：

LongAccumulator maxId = new LongAccumulator(Long::max, 0); //replace 0 with desired initial value
maxId.accumulate(newValue); //from each thread

Answer 3

我相信你所做的是正确的，但我认为这是一个更简单的版本。

private void updateMax(long sample){
      //this takes care of the case where between the comparison and update steps, another thread updates the max

      //For example:
      //if the max value is set to a higher max value than the current value in between the comparison and update step
      //sample will be the higher value from the other thread
      //this means that the sample will now be higher than the current highest (as we just set it to the value passed into this function)
      //on the next iteration of the while loop, we will update max to match the true max value
      //we will then fail the while loop check, and be done with trying to update.
      while(sample > max.get()){
          sample = max.getAndSet(sample);  
      }
}

Answer 4

好像你没有选择答案，这是我的：

// while the update appears bigger than the atomic, try to update the atomic.
private void max(AtomicDouble atomicDouble, double update) {
    double expect = atomicDouble.get();
    while (update > expect) {
        atomicDouble.weakCompareAndSet(expect, update);
        expect = atomicDouble.get();
    }
}

它与接受的答案大致相同，但不使用我个人不喜欢的break或while(true)。

编辑：刚刚在java 8中发现了DoubleAccumulator。文档甚至说这是针对像你这样的摘要统计问题：

DoubleAccumulator max = new DoubleAccumulator(Double::max, Double.NEGATIVE_INFINITY);
parallelStream.forEach(max::accumulate);
max.get();

Answer 5

使用Java 8，您可以利用功能接口和简单的lamda表达式来解决这个问题，只需一行而不是循环：

private void updateMax(long sample) {
    max.updateAndGet(curMax -> (sample > curMax) ? sample : curMax);
}

该解决方案使用updateAndGet(LongUnaryOperator)方法。当前值包含在curMax中，并且如果样本值大于当前最大值，则使用条件运算符执行简单测试，将当前最大值替换为样本值。

线程安全实现max

5 个答案: