如何解決簡單的Java示例以14個(gè)線程運(yùn)行。為什么？？

默認(rèn)情況下，我的JVM（1.6.0_26）產(chǎn)生更多線程。大多數(shù)都具有漂亮的描述性名稱，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

顯然，大多數(shù)線程與內(nèi)存處理有關(guān)：有8個(gè)垃圾收集器線程，外加低內(nèi)存檢測器。Finalizer和ReferenceHandler聽起來像他們也參與了內(nèi)存管理。

C2 CompilerThread0/1 幾乎可以肯定，它與即時(shí)編譯有關(guān)。

至于其余線程的確切目的，我不確定。

解決方法

以下簡單的Java代碼：

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個(gè)線程運(yùn)行。我知道在后臺(tái)運(yùn)行一些GC線程，但是其他線程又有什么用？為什么會(huì)有這么多線程？我在使用Java 1.6.0_26的GentooLinux上。使用Eclipse的編譯器或javac進(jìn)行編譯沒有任何區(qū)別（在Eclipse的調(diào)試模式下運(yùn)行它會(huì)增加3個(gè)線程，但這可能是合理的）。