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id="post-info"><h1 class="post-title">Java并发实现原理-多线程基础-JMM与happen-before(6)</h1><div id="post-meta"><div class="meta-firstline"><span class="post-meta-date"><i class="far fa-calendar-alt fa-fw post-meta-icon"></i><span class="post-meta-label">发表于</span><time class="post-meta-date-created" datetime="2022-04-09T09:23:03.000Z" title="发表于 2022-04-09 17:23:03">2022-04-09</time><span class="post-meta-separator">|</span><i class="fas fa-history fa-fw post-meta-icon"></i><span class="post-meta-label">更新于</span><time class="post-meta-date-updated" datetime="2022-04-09T11:15:10.560Z" title="更新于 2022-04-09 19:15:10">2022-04-09</time></span><span class="post-meta-categories"><span class="post-meta-separator">|</span><i class="fas fa-inbox fa-fw post-meta-icon"></i><a class="post-meta-categories" href="/categories/Java%E5%B9%B6%E5%8F%91%E5%AE%9E%E7%8E%B0%E5%8E%9F%E7%90%86/">Java并发实现原理</a></span></div><div class="meta-secondline"><span class="post-meta-separator">|</span><span class="post-meta-wordcount"><i class="far fa-file-word fa-fw post-meta-icon"></i><span class="post-meta-label">字数总计:</span><span class="word-count">883</span><span class="post-meta-separator">|</span><i class="far fa-clock fa-fw post-meta-icon"></i><span class="post-meta-label">阅读时长:</span><span>2分钟</span></span><span class="post-meta-separator">|</span><span class="post-meta-pv-cv" id="" data-flag-title="Java并发实现原理-多线程基础-JMM与happen-before(6)"><i class="far fa-eye fa-fw post-meta-icon"></i><span class="post-meta-label">阅读量:</span><span id="busuanzi_value_page_pv"></span></span></div></div></div><article class="post-content" id="article-container"><h2 id="1-为什么会存在“内存可见性”问题？"><a href="#1-为什么会存在“内存可见性”问题？" class="headerlink" title="1.为什么会存在“内存可见性”问题？"></a>1.为什么会存在“内存可见性”问题？</h2><p>要解释这个问题，就涉及现代CPU的架构。如下是x86架构下CPU缓存的布局，即在一个CPU4核心下，L1、L2、L3三级缓存与主内存的布局。每个核上面有L1、L2缓存，L3为所有核共用。</p>
<p><img src= "data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7" data-lazy-src="https://cdn.jsdelivr.net/gh/Nesive/nesive-butterfly-cdn/image/java_concurrency_principle/x86_cpu_cache_layout.png" alt="Java并发原理-x86架构下CPU缓存布局"></p>
<p>因为存在CPU缓存一致性协议，例如MESI,多个CPU之间的缓存不会出现不同步的问题，不会有“内存可见性”问题。</p>
<p>但是，缓存一致性协议对性能有很大损耗，为了解决这个问题，CPU的设计者们在这个基础上有进行了各种优化。例如，在计算单算和L1之间家林Store Buffer、Load Buffer(还有其他各种Buffer)，如下图所示。</p>
<p>L1、L2、L3和主内存之间是同步的，有缓存一致性协议的保证，但是Store Buffer、Load Buffer和L1之间确实异步的。也就是说，往内存中写入一个变量，这个变量会保存在Store Buffer里面，稍后才异步地写入L1中，同时同步写入主内存中。</p>
<p>注意这里只是简要花了x86的CPU缓存体系，还没有进一步讨论SMP架构和NUMA的区别，还有其他CPU架构体系，例如PowerPC、MIPS、ARM等，不同CPU的缓存体系会有各种差异。</p>
<p><img src= "data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7" data-lazy-src="https://cdn.jsdelivr.net/gh/Nesive/nesive-butterfly-cdn/image/java_concurrency_principle/x86_with_buffer.png" alt="Java并发原理-x86加了Store Buffer和Load Buffer的CPU缓存体系"></p>
<p>但站在操作系统内核的角度，可以统一看待这件事，也就是如下所示的操作系统内核视角下的CPU缓存模型。</p>
<p><img src= "data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7" data-lazy-src="https://cdn.jsdelivr.net/gh/Nesive/nesive-butterfly-cdn/image/java_concurrency_principle/os_cpu_cache_model.png" alt="Java并发原理-操作系统内核视角下的CPU缓存模型"></p>
<p>多CPU、每隔CPU多核心，每隔核心上面可能还有多个硬件线程，对于操作系统来将，就相当于一个个的逻辑CPU。每隔逻辑CPU都有自己的缓存，这些缓存和主内存之间不是完全同步的。</p>
<p>对应到Java里，就是JVM抽象内存模型，如下图所示。</p>
<p>到此为止，介绍了不同CPU架构下复杂的缓存体系，也就回答了为什么会出现“内存可见性”问题。</p>
<p><img src= "data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7" data-lazy-src="https://cdn.jsdelivr.net/gh/Nesive/nesive-butterfly-cdn/image/java_concurrency_principle/jvm_memory_model.png" alt="Java并发原理-操作系统内核视角下的CPU缓存模型"></p>
<h2 id="2-重排序和内存可见性的关系"><a href="#2-重排序和内存可见性的关系" class="headerlink" title="2.重排序和内存可见性的关系"></a>2.重排序和内存可见性的关系</h2><p>Store Buffer的延迟写入是重排序的一种，成为内存重排序(Memory Ordering)。除此以外，还有编译器和CPU的指令重排序。下面对重排序做了一个分类：</p>
<p>1.编译器重排序。对于没有先后依赖关系的语句，编译器可以重新调整语句的执行顺序。</p>
<p>2.CPU指令重排序。在指令级别，让没有依赖关系的多条指令并行。</p>
<p>3.CPU内存重排序。CPU有自己的缓存，指令执行顺序和写入主内存的顺序不完全一致。</p>
<p>在三种重排序中，第三类就是造成“内存可见性”问题的主要原因，下面再举例来进一步说明这个问题，如下所示：</p>
<pre class="line-numbers language-none"><code class="language-none">线程1：
X = 1
a = Y
线程2：
Y = 1
b = X<span aria-hidden="true" class="line-numbers-rows"><span></span><span></span><span></span><span></span><span></span><span></span><span></span></span></code></pre>
<p>假设X、Y是两个全局变量，初始的时候，X=0，Y=0。请问，这两个线程执行完毕之后，a和b的正确结果应该是什么？</p>
<p>很显然，线程1和线程2的执行先后顺序是不确定的，可能顺序执行，也可能交叉执行，最终正确的结果可能是：</p>
<pre class="line-numbers language-none"><code class="language-none">(1) a = 0,b = 1
(2) a = 1,b = 0
(3) a = 1,b = 1<span aria-hidden="true" class="line-numbers-rows"><span></span><span></span><span></span></span></code></pre>
<p>也就是不管谁先谁后，执行结果应该是这三种场景中的一种。但实际可能是 a = 0, b = 0。</p>
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