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。

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)。

Apache Commons用户可以使用ThreadUtils。当前实现使用前面概述的线程组方法。

for (Thread t : ThreadUtils.getAllThreads()) {
      System.out.println(t.getName() + ", " + t.isDaemon());
}

在java控制台中，点击Ctrl-Break。它将列出所有线程以及堆的一些信息。当然，这不会让您访问对象。但无论如何，它对调试都很有帮助。

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。

    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 + "  ");
        }
    }