简单的Java示例以14个线程运行。为什么？

如何解决简单的Java示例以14个线程运行。为什么？？

默认情况下，我的JVM（1.6.0_26）产生更多线程。大多数都具有漂亮的描述性名称，以暗示其目的：

@H_404_2@'Attach Listener' daemon prio=10 tid=0x0000000041426800 nid=0x2fb9 waiting on condition [0x0000000000000000] java.lang.Thread.State: RUNNABLE'Low Memory Detector' daemon prio=10 tid=0x00007f512c07e800 nid=0x2fa3 runnable [0x0000000000000000] java.lang.Thread.State: RUNNABLE'C2 CompilerThread1' daemon prio=10 tid=0x00007f512c07b800 nid=0x2fa2 waiting on condition [0x0000000000000000] java.lang.Thread.State: RUNNABLE'C2 CompilerThread0' daemon prio=10 tid=0x00007f512c078800 nid=0x2fa1 waiting on condition [0x0000000000000000] java.lang.Thread.State: RUNNABLE'Signal dispatcher' daemon prio=10 tid=0x00007f512c076800 nid=0x2fa0 runnable [0x0000000000000000] java.lang.Thread.State: RUNNABLE'Finalizer' daemon prio=10 tid=0x00007f512c05a000 nid=0x2f9f in Object.wait() [0x00007f512b8f7000] java.lang.Thread.State: WAITING (on object monitor) at java.lang.Object.wait(Native Method) - waiting on <0x00000007c14b1300> (a java.lang.ref.ReferenceQueue$Lock) at java.lang.ref.ReferenceQueue.remove(ReferenceQueue.java:118) - locked <0x00000007c14b1300> (a java.lang.ref.ReferenceQueue$Lock) at java.lang.ref.ReferenceQueue.remove(ReferenceQueue.java:134) at java.lang.ref.Finalizer$FinalizerThread.run(Finalizer.java:159)'Reference Handler' daemon prio=10 tid=0x00007f512c058000 nid=0x2f9e in Object.wait() [0x00007f512b9f8000] java.lang.Thread.State: WAITING (on object monitor) at java.lang.Object.wait(Native Method) - waiting on <0x00000007c14b11d8> (a java.lang.ref.Reference$Lock) at java.lang.Object.wait(Object.java:485) at java.lang.ref.Reference$ReferenceHandler.run(Reference.java:116) - locked <0x00000007c14b11d8> (a java.lang.ref.Reference$Lock)'main' prio=10 tid=0x0000000041401800 nid=0x2f94 waiting on condition [0x00007f5135735000] java.lang.Thread.State: TIMED_WAITING (sleeping) at java.lang.Thread.sleep(Native Method) at Main.main(Main.java:5)'VM Thread' prio=10 tid=0x00007f512c051800 nid=0x2f9d runnable'GC task thread#0 (ParallelGC)' prio=10 tid=0x0000000041414800 nid=0x2f95 runnable'GC task thread#1 (ParallelGC)' prio=10 tid=0x00007f512c001000 nid=0x2f96 runnable'GC task thread#2 (ParallelGC)' prio=10 tid=0x00007f512c002800 nid=0x2f97 runnable'GC task thread#3 (ParallelGC)' prio=10 tid=0x00007f512c004800 nid=0x2f98 runnable'GC task thread#4 (ParallelGC)' prio=10 tid=0x00007f512c006800 nid=0x2f99 runnable'GC task thread#5 (ParallelGC)' prio=10 tid=0x00007f512c008000 nid=0x2f9a runnable'GC task thread#6 (ParallelGC)' prio=10 tid=0x00007f512c00a000 nid=0x2f9b runnable'GC task thread#7 (ParallelGC)' prio=10 tid=0x00007f512c00c000 nid=0x2f9c runnable'VM Periodic Task Thread' prio=10 tid=0x00007f512c089000 nid=0x2fa4 waiting on condition

显然，大多数线程与内存处理有关：有8个垃圾收集器线程，外加低内存检测器。Finalizer和ReferenceHandler听起来像他们也参与了内存管理。

C2 CompilerThread0/1 几乎可以肯定，它与即时编译有关。

至于其余线程的确切目的，我不确定。

解决方法

以下简单的Java代码：

public class Main { public static void main(String[] args) throws InterruptedException {System.out.println('Start');Thread.sleep(5000);System.out.println('Done'); }}

使用14个线程运行。我知道在后台运行一些GC线程，但是其他线程又有什么用？为什么会有这么多线程？我在使用Java 1.6.0_26的GentooLinux上。使用Eclipse的编译器或javac进行编译没有任何区别（在Eclipse的调试模式下运行它会增加3个线程，但这可能是合理的）。