小弟最近在翻看JDK的源碼但是奈何基礎(chǔ)太差有一處始終想不通。希望大家?guī)兔獯鹣鹿餐M步。ConcurrentLinkedQueue的源碼中的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; 那么對p的操作應(yīng)該等同與對t的操作，那么將newNode設(shè)置為p.next不久等同于對t也設(shè)置next屬性么， 為什么p！=t呢？
2 回答
侃侃爾雅
TA貢獻1801條經(jīng)驗 獲得超16個贊
就一個線程在處理的話肯定不會走casTail這句，如果有多個線程呢？
public boolean offer(E e) {
checkNotNull(e);
final Node<E> newNode = new Node<E>(e);
for (Node<E> t = tail, p = t;;) {
//如果2個線程到這里，一個先把流程走完了
//第二個線程的q開始執(zhí)行的時候已經(jīng)不是null了
//所以會走else的分支，修改了p和t的關(guān)系
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
}
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;
}
}
ibeautiful
TA貢獻1993條經(jīng)驗 獲得超6個贊
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