const{createApp}=Vue const{createVuetify,useGoTo,useDisplay}=Vuetify var data={alert:{show:false,color:'success',text:'',timeout:0,},theme:{dark:false,},nav:{showDrawer:false,showTOC:true,tocPanel:0,tab:'account',post:{discussionId:1005,currentPage:1,targetPage:1,total:8,anchor:0,goToOptions:{container:null,duration:0,easing:'easeInOutCubic',offset:-100,},worker:null,task:[],active:[],apiLock:[],originLike:new Map([]),},related:{block:1,}},search:{width:80,text:null,loading:false,},tags:[{id:28,url:'/t/3d104b0f6a045c7568772c310313',name:'Java',color:'#FFA726',icon:'mdi-language-java',},],posts:[{id:12390,num:0,uid:4645,content:'背景\u003Cp\u003E偶然看到了一篇 ReentrantLock 的源码分析文章，自己便去学习了一下源码。核心思想是如果被请求的共享资源空闲，那么就将当前请求资源的线程设置为有效的工作线程，将共享资源设置为锁定状态；如果共享资源被占用，就需要一定的阻塞等待唤醒机制来保证锁分配。\u003C/p\u003E问题\u003Cp\u003EReentrantLock 的 lock 方法在入队之前直接获取锁，是否冗余设计，存在冗余的代码执行？\u003C/p\u003E我的观点\u003Cp\u003E进行了冗余设计，理解如下。\u003C/p\u003Ejdk8\u003Cp\u003Elock 简化代码如下，我的观点是：非公平锁(NofairSync) lock 方法，有两次尝试直接获取锁，是否冗余设计了？\u003C/p\u003E\u003Ccode\u003Estatic final class NonfairSync extends Sync { final void lock() { // 直接尝试获取锁，不公平表现 if (compareAndSetState(0, 1)) setExclusiveOwnerThread(Thread.currentThread()); else acquire(1); // acquire 方法中首先会调用 tryAcquire ，会第二次尝试获取锁 } protected final boolean tryAcquire(int acquires) { // 如果资源空闲，直接尝试获取锁 // 如果资源不空闲，判断是否当前线程占有，进行重入锁 }}static final class FairSync extends Sync { final void lock() { acquire(1); } protected final boolean tryAcquire(int acquires) { // 如果资源空闲，判断是否已经入队，进行公平处理 // 如果资源不空闲，判断是否当前线程占有，进行重入锁 }}class AbstractQueuedSynchronizer { public final void acquire(int arg) { if (!tryAcquire(arg) \u0026amp;\u0026amp; acquireQueued(addWaiter(Node.EXCLUSIVE), arg)) selfInterrupt(); }}\u003C/code\u003Ejdk17\u003Cp\u003Elock 简化代码如下，我的观点是：lock 方法，有两次尝试直接获取锁，是否冗余设计了？对于公平锁，均会判断当前线程是否已经入队，是冗余的尝试；对于非公平锁，连续两次尝试直接获取锁，也是冗余的尝试\u003C/p\u003E\u003Ccode\u003Eabstract static class Sync extends AbstractQueuedSynchronizer { abstract boolean initialTryLock(); final void lock() { if (!initialTryLock()) // 首次尝试获取锁 acquire(1); // acquire 方法中首先会调用 tryAcquire ，会第二次尝试获取锁 }}static final class NonfairSync extends Sync { final boolean initialTryLock() { // 如果资源空闲，直接尝试获取锁 // 如果资源不空闲，判断是否当前线程占有，进行重入锁 } protected final boolean tryAcquire(int acquires) { // 如果资源空闲，直接尝试获取锁 }}static final class FairSync extends Sync { final boolean initialTryLock() { // 如果资源空闲，判断是否已经入队，进行公平处理 // 如果资源不空闲，判断是否当前线程占有，进行重入锁 } protected final boolean tryAcquire(int acquires) { // 如果资源空闲，判断是否已经入队，进行公平处理 }}\u003C/code\u003E我认为的不冗余的方式\u003Ccode\u003Estatic class Sync extends AbstractQueuedSynchronizer { final void lock() { acquire(1); }}static final class NonfairSync extends Sync { protected final boolean tryAcquire(int acquires) { // 如果资源空闲，直接尝试获取锁 // 如果资源不空闲，判断是否当前线程占有，进行重入锁 }}static final class FairSync extends Sync { protected final boolean tryAcquire(int acquires) { // 如果资源空闲，判断是否已经入队，进行公平处理 // 如果资源不空闲，判断是否当前线程占有，进行重入锁 }}\u003C/code\u003E',ipRegion:'',updatedByUid:0,createdAt:'2025-03-10 12:33:59',updatedAt:'2025-03-12 13:23:13',mentionNum:0,mentionedBy:[],mentionUsers:[],likeUsers:[],},{id:12391,num:1,uid:2671,content:'不算设计冗余，两次调用 compareAndSetState 都是为了在无竞争时快速成功。第一次调用返回 false 意味着有竞争，此时进入 acquire 。在 acquire 里会根据 state 的值判断是否调用 compareAndSetState ，原因是这中间有可能从有竞争变为无竞争（ volatile ），此时快速成功就行了。state 的值\u0026gt;0 才是可重入锁/其它线程占用的情况，最后再做其它处理。',ipRegion:'',updatedByUid:0,createdAt:'2025-03-10 14:52:02',updatedAt:'2025-03-12 13:23:13',mentionNum:0,mentionedBy:[],mentionUsers:[],likeUsers:[],},{id:12392,num:2,uid:472,content:'你的疑问算是一个面试题了',ipRegion:'',updatedByUid:0,createdAt:'2025-03-10 14:55:23',updatedAt:'2025-03-12 13:23:13',mentionNum:0,mentionedBy:[],mentionUsers:[],likeUsers:[],},{id:12393,num:3,uid:223,content:'经典面试题。跟一楼说的一样，大部分时候资源是不上锁的，一般能快速拿到锁就拉倒了没必要再去 AQS 里面拐一圈。\u003Cbr\u003E这也算是并发编程里面的“快慢路径”的一种体现。',ipRegion:'',updatedByUid:0,createdAt:'2025-03-10 15:47:16',updatedAt:'2025-03-12 13:23:13',mentionNum:0,mentionedBy:[],mentionUsers:[],likeUsers:[],},{id:12394,num:4,uid:4646,content:'学到了',ipRegion:'',updatedByUid:0,createdAt:'2025-03-10 15:54:00',updatedAt:'2025-03-12 13:23:13',mentionNum:0,mentionedBy:[],mentionUsers:[],likeUsers:[],},{id:12395,num:5,uid:4647,content:'其实把你的问题直接问 AI ，都有答案了。\u003Cbr\u003E1. 非公平锁的“两次尝试”设计\u003Cbr\u003E在非公平锁中，lock() 方法在入队前先尝试直接获取锁（通过 initialTryLock() 或 CAS 操作），若失败再调用 acquire()，而在 acquire() 中又会调用 tryAcquire() 再次尝试。这看似重复，实为优化：\u003Cbr\u003E\u003Cbr\u003E减少线程挂起/唤醒的开销：当锁被释放时，新线程可能直接抢占锁（插队），而无需等待队列中的线程被唤醒，提高了吞吐量。\u003Cbr\u003E避免不必要的队列操作：若首次尝试成功，线程直接获取锁，无需进入队列，减少了入队、出队和上下文切换的开销。',ipRegion:'',updatedByUid:0,createdAt:'2025-03-10 16:19:40',updatedAt:'2025-03-12 13:23:13',mentionNum:0,mentionedBy:[],mentionUsers:[],likeUsers:[],},{id:12396,num:6,uid:4648,content:'并发编程的经典优化方式：快慢路径\u003Cbr\u003E\u003Cbr\u003E一般先走 Fast Path ，用 CAS 尝试快速获取资源；失败后，走 Slow Path ，可能会涉及到一些阻塞、沉睡、排队之类的操作',ipRegion:'',updatedByUid:0,createdAt:'2025-03-10 20:18:07',updatedAt:'2025-03-12 13:23:13',mentionNum:0,mentionedBy:[],mentionUsers:[],likeUsers:[],},{id:12397,num:7,uid:1131,content:'我觉得要重点提出 1 楼说的“这中间有可能从有竞争变为无竞争（ volatile ）”，我觉了楼主可能犯了并发编程里一个常见是思维错误就是我“刚刚” 判断了一个条件，后面就不用判断了。',ipRegion:'',updatedByUid:0,createdAt:'2025-03-11 00:05:19',updatedAt:'2025-03-12 13:23:13',mentionNum:0,mentionedBy:[],mentionUsers:[],likeUsers:[],},],usersMap:new 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