如何显示，双重检查锁图案与字典的TryGetValue不是线程 [英] How to show that the double-checked-lock pattern with Dictionary's TryGetValue is not threadsafe

问题描述

最近我已经看到了使用双检锁模式在词典一些C＃项目。事情是这样的：

Recently I've seen some C# projects that use a double-checked-lock pattern on a Dictionary. Something like this:

private static readonly object _lock = new object();
private static volatile IDictionary<string, object> _cache = 
    new Dictionary<string, object>();

public static object Create(string key)
{
    object val;
    if (!_cache.TryGetValue(key, out val))
    {
        lock (_lock)
        {
            if (!_cache.TryGetValue(key, out val))
            {
                val = new object(); // factory construction based on key here.
                _cache.Add(key, val);
            }
        }
    }
    return val;
}

这code是不正确的，因为在词典可以生长在 _cache.Add集合（），而 _cache.TryGetValue （锁外）被遍历集合。这可能是在许多情况的可能性微乎其微，但仍然是错误的。

This code is incorrect, since the Dictionary can be "growing" the collection in _cache.Add() while _cache.TryGetValue (outside the lock) is iterating over the collection. It might be extremely unlikely in many situations, but is still wrong.

有一个简单的程序来演示这个code失败？

Is there a simple program to demonstrate that this code fails?

是否有意义将其纳入一个单元测试？如果是的话，怎么办？

Does it make sense to incorporate this into a unit test? And if so, how?

推荐答案

在这个例子中，异常＃1几乎是瞬间在我的机器上抛出：

In this example, exception #1 is thrown almost instantly on my machine:

var dict = new Dictionary<int, string>() { { 1234, "OK" } };

new Thread(() =>
{
    for (; ; )
    {
        string s;
        if (!dict.TryGetValue(1234, out s))
        {
            throw new Exception();  // #1
        }
        else if (s != "OK")
        {
            throw new Exception();  // #2
        }
    }
}).Start();

Thread.Sleep(1000);
Random r = new Random();
for (; ; )
{
    int k;
    do { k = r.Next(); } while (k == 1234);
    Debug.Assert(k != 1234);
    dict[k] = "FAIL";
}

不过，code，它不是设计的确切行为是线程安全的是未predictable 。
您不能依靠它。因此，仔细检查code确实是坏了。

However, the exact behaviour of code that is not designed to be thread-safe is unpredictable.
You cannot rely on it. So the double-checking code is indeed broken.

我不知道如果我的单元测试这一点，虽然，作为测试并发code（和获得它的权利）要复杂得多比写并发code摆在首位。

I'm not sure if I'd unit test this, though, as testing concurrent code (and getting it right) is much more complicated than writing the concurrent code in the first place.

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