最近在翻看JDK的源码但是奈何基础太差有一处始终想不通。希望大家帮忙解答下共同进步。
1
724ConcurrentLinkedQueue的源码中的offer方法
（checkNotNull(e);
final Node<E> newNode = new Node<E>(e);
for (Node<E> t = tail, p = t;;) {
Node<E> q = p.next;
if (q == null) {
// p is last node
if (p.casNext(null, newNode)) {
// Successful CAS is the linearization point
// for e to become an element of this queue,
// and for newNode to become "live".
if (p != t) // hop two nodes at a time
casTail(t, newNode); // Failure is OK.
return true;
}
// Lost CAS race to another thread; re-read next
}
）我的理解是 既然t = p; 那么对p的操作应该等同与对t的操作，那么将newNode设置为p.next不久等同于对t也设置next属性么， 为什么p！=t呢？
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就一个线程在处理的话肯定不会走casTail这句，如果有多个线程呢？
public boolean offer(E e) { checkNotNull(e); final Node newNode = new Node(e);
for (Node t = tail, p = t;;) {
//如果2个线程到这里，一个先把流程走完了
//第二个线程的q开始执行的时候已经不是null了
//所以会走else的分支，修改了p和t的关系
Node q = p.next;
if (q == null) {
// p is last node
if (p.casNext(null, newNode)) {
// Successful CAS is the linearization point
// for e to become an element of this queue,
// and for newNode to become "live".
if (p != t) // hop two nodes at a time
casTail(t, newNode); // Failure is OK.
return true;
}
// Lost CAS race to another thread; re-read next
}
else if (p == q)
// We have fallen off list. If tail is unchanged, it
// will also be off-list, in which case we need to
// jump to head, from which all live nodes are always
// reachable. Else the new tail is a better bet.
p = (t != (t = tail)) ? t : head;
else
// Check for tail updates after two hops.
p = (p != t && t != (t = tail)) ? t : q;
}
}