java – 线程安全的排序链表

细粒度：

public void add(T t) {                                                         
  Node curr = head;
  curr.lock.lock();

  while (curr.next != null) {
    // Invariant: curr is locked                                               
    // Invariant: curr.data < t                                                
    curr.next.lock.lock();                                                     

    if (t.compareTo(curr.next.data) <= 0) {                                    
      break;                                                                   
    }                                                                          

    Node tmp = curr.next;                                                      
    curr.lock.unlock();                                                        
    curr = tmp;                                                                
  }                                                                            

  // curr is acquired                                                          
  curr.next = new Node(curr.next,t);                                          
  if (curr.next.next != null) {  // old curr's next is acquired                
    curr.next.next.lock.unlock();                                              
  }                                                                            
  curr.lock.unlock();                                                          
}

粗粒度：

public void add(T t) {
  lock.lock();
  Node curr = head;
  while (curr.next != null) {
    if (t.compareTo(curr.next.data) <= 0) {
      break;
    }                                                                          
    curr = curr.next;                                                          
  }                                                                            
  curr.next = new Node(curr.next,t);                                          
  lock.unlock();                                                               
}

我将4个线程(在4个逻辑CPU核心上)的两个版本定时插入20000个整数.每个线程的时间显示CPU时间(即它不包括等待时间).

Fine grained:
Worked 1 spent 1080 ms
Worked 2 spent 1230 ms
Worked 0 spent 1250 ms
Worked 3 spent 1260 ms
wall time: 1620 ms

Coarse grained:
Worked 1 spent 190 ms
Worked 2 spent 270 ms
Worked 3 spent 410 ms
Worked 0 spent 280 ms
wall time: 1298 ms

我最初的想法是.lock()和.unlock()是问题,但我分析了实现,他们一起只消耗了30％的时间.我的第二个猜测是,细粒度的解决方案有更多的缓存未命中,但我怀疑它,因为单个链表与数组不同,本质上容易出现缓存未命中.

知道为什么我没有得到预期的并行化吗？