获取两个互斥锁的锁，避免死锁 [英] Acquire a lock on two mutexes and avoid deadlock

问题描述

以下代码包含潜在的死锁，但似乎是必要的：要将数据安全地从另一个容器复制到一个容器，必须锁定这两个容器，以防止在另一个线程中发生更改。

The following code contains a potential deadlock, but seems to be necessary: to safely copy data to one container from another, both containers must be locked to prevent changes from occurring in another thread.

void foo::copy(const foo & rhs)
{
    pMutex->lock();
    rhs.pMutex->lock();
    // do copy
}

Foo有一个STL容器，复制基本上由使用std :: copy组成。

Foo has an STL container and "do copy" essentially consists of using std::copy. How do I lock both mutexes without introducing deadlock?

推荐答案

对的实例施加某种总排序， foo 并始终以递增或递减顺序获取它们的锁，例如， foo1-> lock（）然后 foo2-> lock（）。

Impose some kind of total order on instances of foo and always acquire their locks in either increasing or decreasing order, e.g., foo1->lock() and then foo2->lock().

另一种方法是使用函数语义， c $ c> foo :: clone 方法创建一个新的实例，而不是破坏现有的实例。

Another approach is to use functional semantics and instead write a foo::clone method that creates a new instance rather than clobbering an existing one.

如果你的代码做了很多锁定，您可能需要一个复杂的死锁避免算法，例如银行家算法。

If your code is doing lots of locking, you may need a complex deadlock-avoidance algorithm such as the banker's algorithm.

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