在Java中，如果並發執行`insert（）`和`size（）`是否會有死鎖？

Question

代碼看起來像這樣（ link ）：

/***
 * Excerpted from "Seven Concurrency Models in Seven Weeks",
***/
import java.util.concurrent.locks.ReentrantLock;

class ConcurrentSortedList {

  private class Node {
    int value;
    Node prev;
    Node next;
    ReentrantLock lock = new ReentrantLock();

    Node() {}

    Node(int value, Node prev, Node next) {
      this.value = value; this.prev = prev; this.next = next;
    }
  }

  private final Node head;
  private final Node tail;

  public ConcurrentSortedList() {
    head = new Node(); tail = new Node();
    head.next = tail; tail.prev = head;
  }

  public void insert(int value) {
    Node current = head;
    current.lock.lock(); 
    Node next = current.next;
    try {
      while (true) {
        next.lock.lock(); 
        try {
          if (next == tail || next.value < value) { 
            Node node = new Node(value, current, next); 
            next.prev = node;
            current.next = node;
            return; 
          }
        } finally { current.lock.unlock(); } 
        current = next;
        next = current.next;
      }
    } finally { next.lock.unlock(); } 
  }

  public int size() {
    Node current = tail;
    int count = 0;

    while (current.prev != head) {
      ReentrantLock lock = current.lock;
      lock.lock();
      try {
        ++count;
        current = current.prev;
      } finally { lock.unlock(); }
    }

    return count;
  }

}

它說它使用了交接鎖定 。 insert()從列表頭到列表尾獲取鎖定， size()從列表頭到列表頭部獲取鎖定。 size()和insert()可以並發執行。

但是我認為size()和insert()不能並發執行。 因為如果insert持有aNode上的鎖並請求aNode.next上的鎖，而size持有aNode.next上的鎖並請求aNode上的鎖，則將出現死鎖。

有人對此有想法嗎？ 謝謝！

Answer 1

我看到.. size()將在請求新鎖之前釋放當前鎖。.因此不會出現死鎖。