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JavaScript闭包如何被垃圾回收(How JavaScript closures are garbage collected)

 我已经记录了以下Chrome错误 ，这导致我的代码中出现了许多严重和非明显的内存泄漏：  
 （这些结果使用Chrome开发工具的内存分析器 ，它运行GC，然后对不收集垃圾的所有内容进行堆快照。）  
 在下面的代码中， someClass实例是垃圾回收（不错）：  
var someClass = function() {};

function f() {
  var some = new someClass();
  return function() {};
}

window.f_ = f();
 
 但在这种情况下不会收集垃圾（不好）：  
var someClass = function() {};

function f() {
  var some = new someClass();
  function unreachable() { some; }
  return function() {};
}

window.f_ = f();
 
 和相应的截图：  
 
 似乎一个闭包（在这种情况下， function() {} ）保持所有对象“活着”，如果该对象被同一上下文中的任何其他闭包引用，无论该闭包本身甚至是否可达。  
 我的问题是在其他浏览器（IE 9+和Firefox）中关闭垃圾收集。 我非常熟悉webkit的工具，例如JavaScript堆分析器，但是我不了解其他浏览器的工具，所以我无法测试。  
 在这三种情况中，哪些IE9 +和Firefox垃圾收集 someClass 实例？ 

I've logged the following Chrome bug, which has led to many serious and non-obvious memory leaks in my code: 
(These results use Chrome Dev Tools' memory profiler, which runs the GC, and then takes a heap snapshot of everything not garbaged collected.) 
In the code below, the someClass instance is garbage collected (good): 
var someClass = function() {};

function f() {
  var some = new someClass();
  return function() {};
}

window.f_ = f();
 
But it won't be garbage collected in this case (bad): 
var someClass = function() {};

function f() {
  var some = new someClass();
  function unreachable() { some; }
  return function() {};
}

window.f_ = f();
 
And the corresponding screenshot: 
 
It seems that a closure (in this case, function() {}) keeps all objects "alive" if the object is referenced by any other closure in the same context, whether or not if that closure itself is even reachable. 
My question is about garbage collection of closure in other browsers (IE 9+ and Firefox). I am quite familiar with webkit's tools, such as the JavaScript heap profiler, but I know little of other browsers' tools, so I haven't been able to test this. 
In which of these three cases will IE9+ and Firefox garbage collect the someClass instance?

原文：https://stackoverflow.com/questions/19798803

更新时间：2023-07-22 12:07

最满意答案

 从InnoDB中删除数据是您可以请求的最昂贵的操作。 正如您已经发现查询本身不是问题 - 大多数都将针对相同的执行计划进行优化。  
 虽然可能很难理解为什么所有案例的删除都是最慢的，但是有一个相当简单的解释。 InnoDB是一个事务性存储引擎。 这意味着如果您的查询中途中止，则所有记录仍将保留，就好像没有发生任何事情。 一旦完成，所有的都将在同一时刻消失。 在DELETE期间，连接到服务器的其他客户端将看到记录，直到您的DELETE完成。  
 为了实现这一点，InnoDB使用了一种名为MVCC（Multi Version Concurrency Control）的技术。 它基本上做的是为每个连接提供整个数据库的快照视图，就像在事务的第一个语句开始时一样。 为了实现这一点，InnoDB内的每个记录都可以有多个值 - 每个快照一个。 这也是为什么InnoDB上的COUNTing需要一些时间 - 这取决于当时看到的快照状态。  
 对于您的DELETE事务，根据查询条件识别的每个记录都被标记为删除。 由于其他客户端可能同时访问数据，因此无法立即将其从表中删除，因为它们必须查看其各自的快照以保证删除的原子性。  
 一旦所有记录都被标记为删除，则事务被成功提交。 即使这样，在DELETE事务之前所有其他使用快照值的事务也已经结束的情况下，它们不能立即从实际数据页中删除。  
 所以实际上你的3分钟并不是那么慢，考虑到所有的记录必须被修改，以便以交易安全的方式来准备它们去除。 可能您会在声明运行时“听到”您的硬盘工作。 这是由于访问所有行。 为了提高性能，您可以尝试增加服务器的InnoDB缓冲池大小，并尝试在DELETE时限制对数据库的其他访问，从而也减少InnoDB每个记录维护的历史版本数。 随着额外的内存，InnoDB可能能够将表（主要）读入内存，并避免了一些磁盘寻找时间。 

Deleting data from InnoDB is the most expensive operation you can request of it. As you already discovered the query itself is not the problem - most of them will be optimized to the same execution plan anyway. 
While it may be hard to understand why DELETEs of all cases are the slowest, there is a rather simple explanation. InnoDB is a transactional storage engine. That means that if your query was aborted halfway-through, all records would still be in place as if nothing happened. Once it is complete, all will be gone in the same instant. During the DELETE other clients connecting to the server will see the records until your DELETE completes. 
To achieve this, InnoDB uses a technique called MVCC (Multi Version Concurrency Control). What it basically does is to give each connection a snapshot view of the whole database as it was when the first statement of the transaction started. To achieve this, every record in InnoDB internally can have multiple values - one for each snapshot. This is also why COUNTing on InnoDB takes some time - it depends on the snapshot state you see at that time. 
For your DELETE transaction, each and every record that is identified according to your query conditions, gets marked for deletion. As other clients might be accessing the data at the same time, it cannot immediately remove them from the table, because they have to see their respective snapshot to guarantee the atomicity of the deletion. 
Once all records have been marked for deletion, the transaction is successfully committed. And even then they cannot be immediately removed from the actual data pages, before all other transactions that worked with a snapshot value before your DELETE transaction, have ended as well.  
So in fact your 3 minutes are not really that slow, considering the fact that all records have to be modified in order to prepare them for removal in a transaction safe way. Probably you will "hear" your hard disk working while the statement runs. This is caused by accessing all the rows. To improve performance you can try to increase InnoDB buffer pool size for your server and try to limit other access to the database while you DELETE, thereby also reducing the number of historic versions InnoDB has to maintain per record. With the additional memory InnoDB might be able to read your table (mostly) into memory and avoid some disk seeking time.

JavaScript闭包如何被垃圾回收(How JavaScript closures are garbage collected)

最满意答案

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