我想知道我的线程可以运行的最佳数量。 通常情况下，这相当于Runtime.getRuntime().availableProcessors()
但是，返回的次数为两次一个CPU支持超线程的那样高。 现在，对于某些任务超线程是好的，但对其他人而言，什么都不做。 就我而言，我怀疑，它什么都不做，所以我想知道我是否有除以返回的数字Runtime.getRuntime().availableProcessors()两种。
对于我推断出CPU是否支持超线程。 因此，我的问题 - 我怎么能做到这一点在Java中？
谢谢。
编辑
OK，我已经基准我的代码。 下面是我的环境：
- 联想ThinkPad W510（即i7的CPU有4个内核和超线程）的RAM，16G
- Windows 7的
- 84个压缩的CSV文件有拉链尺寸范围从105M 16M的
- 所有文件都逐个读取在主线程中 - 没有多线程访问HD。
- 每个CSV文件行包含一些数据，这些数据被分析和快速上下文测试确定该行是否是相关的。
- 每个相关的行包含两个双打（表示经度和纬度，为好奇），其被强制转换为单个
Long，然后将其存储在共享哈希集合。
因此，工作线程不读从HD什么，但他们不与解压缩和解析的内容（使用占用自己opencsv库）。
下面是代码，W / O无聊的细节：
public void work(File dir) throws IOException, InterruptedException {
Set<Long> allCoordinates = Collections.newSetFromMap(new ConcurrentHashMap<Long, Boolean>());
int n = 6;
// NO WAITING QUEUE !
ThreadPoolExecutor exec = new ThreadPoolExecutor(n, n, 0L, TimeUnit.MILLISECONDS, new SynchronousQueue<Runnable>());
StopWatch sw1 = new StopWatch();
StopWatch sw2 = new StopWatch();
sw1.start();
sw2.start();
sw2.suspend();
for (WorkItem wi : m_workItems) {
for (File file : dir.listFiles(wi.fileNameFilter)) {
MyTask task;
try {
sw2.resume();
// The only reading from the HD occurs here:
task = new MyTask(file, m_coordinateCollector, allCoordinates, wi.headerClass, wi.rowClass);
sw2.suspend();
} catch (IOException exc) {
System.err.println(String.format("Failed to read %s - %s", file.getName(), exc.getMessage()));
continue;
}
boolean retry = true;
while (retry) {
int count = exec.getActiveCount();
try {
// Fails if the maximum of the worker threads was created and all are busy.
// This prevents us from loading all the files in memory and getting the OOM exception.
exec.submit(task);
retry = false;
} catch (RejectedExecutionException exc) {
// Wait for any worker thread to finish
while (exec.getActiveCount() == count) {
Thread.sleep(100);
}
}
}
}
}
exec.shutdown();
exec.awaitTermination(1, TimeUnit.HOURS);
sw1.stop();
sw2.stop();
System.out.println(String.format("Max concurrent threads = %d", n));
System.out.println(String.format("Total file count = %d", m_stats.getFileCount()));
System.out.println(String.format("Total lines = %d", m_stats.getTotalLineCount()));
System.out.println(String.format("Total good lines = %d", m_stats.getGoodLineCount()));
System.out.println(String.format("Total coordinates = %d", allCoordinates.size()));
System.out.println(String.format("Overall elapsed time = %d sec, excluding I/O = %d sec", sw1.getTime() / 1000, (sw1.getTime() - sw2.getTime()) / 1000));
}
public class MyTask<H extends CsvFileHeader, R extends CsvFileRow<H>> implements Runnable {
private final byte[] m_buffer;
private final String m_name;
private final CoordinateCollector m_coordinateCollector;
private final Set<Long> m_allCoordinates;
private final Class<H> m_headerClass;
private final Class<R> m_rowClass;
public MyTask(File file, CoordinateCollector coordinateCollector, Set<Long> allCoordinates,
Class<H> headerClass, Class<R> rowClass) throws IOException {
m_coordinateCollector = coordinateCollector;
m_allCoordinates = allCoordinates;
m_headerClass = headerClass;
m_rowClass = rowClass;
m_name = file.getName();
m_buffer = Files.toByteArray(file);
}
@Override
public void run() {
try {
m_coordinateCollector.collect(m_name, m_buffer, m_allCoordinates, m_headerClass, m_rowClass);
} catch (IOException e) {
e.printStackTrace(); //To change body of catch statement use File | Settings | File Templates.
}
}
}
请在下面找到结果（我稍微改变了输出省略重复部分）：
Max concurrent threads = 4
Total file count = 84
Total lines = 56395333
Total good lines = 35119231
Total coordinates = 987045
Overall elapsed time = 274 sec, excluding I/O = 266 sec
Max concurrent threads = 6
Overall elapsed time = 218 sec, excluding I/O = 209 sec
Max concurrent threads = 7
Overall elapsed time = 209 sec, excluding I/O = 199 sec
Max concurrent threads = 8
Overall elapsed time = 201 sec, excluding I/O = 192 sec
Max concurrent threads = 9
Overall elapsed time = 198 sec, excluding I/O = 186 sec
你可以自由地得出自己的结论，但我就是不超线程提高我的具体情况下的性能。 此外，具有6个工作线程似乎是这个任务，我的机器是正确的选择。
