获取一个可迭代集:

Set<Thread> threadSet = Thread.getAllStackTraces().keySet();

性能:0毫秒，12个线程(Azul JVM 16.0.1, Windows 10, Ryzen 5600X)。

获取根线程组的句柄，像这样:

ThreadGroup rootGroup = Thread.currentThread().getThreadGroup();
ThreadGroup parentGroup;
while ((parentGroup = rootGroup.getParent()) != null) {
    rootGroup = parentGroup;
}

现在，在根组上重复调用enumerate()函数。第二个参数让你递归地获取所有线程:

Thread[] threads = new Thread[rootGroup.activeCount()];
while (rootGroup.enumerate(threads, true ) == threads.length) {
    threads = new Thread[threads.length * 2];
}

注意我们如何反复调用enumerate()，直到数组大到足以包含所有条目。

你可以尝试这样做:

Thread.getAllStackTraces().keySet().forEach((t) -> System.out.println(t.getName() + "\nIs Daemon " + t.isDaemon() + "\nIs Alive " + t.isAlive()));

如果你需要，你可以得到更多的线程特性。

您可以使用返回所有活动线程id的getAllThreadIds。当此方法返回时，返回数组中包含的某些线程可能已经终止。

ManagementFactory.getThreadMXBean().getAllThreadIds()

您查看过jconsole吗?

这将列出为特定Java进程运行的所有线程。

可以从JDK bin文件夹启动jconsole。

你也可以通过在Windows中按Ctrl+Break或在Linux中发送kill pid——QUIT来获得所有线程的完整堆栈跟踪。

    public static void main(String[] args) {


        // Walk up all the way to the root thread group
        ThreadGroup rootGroup = Thread.currentThread().getThreadGroup();
        ThreadGroup parent;
        while ((parent = rootGroup.getParent()) != null) {
            rootGroup = parent;
        }

        listThreads(rootGroup, "");
    }


    // List all threads and recursively list all subgroup
    public static void listThreads(ThreadGroup group, String indent) {
        System.out.println(indent + "Group[" + group.getName() + 
                ":" + group.getClass()+"]");
        int nt = group.activeCount();
        Thread[] threads = new Thread[nt*2 + 10]; //nt is not accurate
        nt = group.enumerate(threads, false);

        // List every thread in the group
        for (int i=0; i<nt; i++) {
            Thread t = threads[i];
            System.out.println(indent + "  Thread[" + t.getName() 
                    + ":" + t.getClass() + "]");
        }

        // Recursively list all subgroups
        int ng = group.activeGroupCount();
        ThreadGroup[] groups = new ThreadGroup[ng*2 + 10];
        ng = group.enumerate(groups, false);

        for (int i=0; i<ng; i++) {
            listThreads(groups[i], indent + "  ");
        }
    }