安全的方式从多线程获取单例实例 [英] safe way to get a singleton instance from multithreads

问题描述

方法1

DataCenter* DataCenter::getInstance()
{
    static DataCenter instance;
    return &instance;    
}

方法2 b

DataCenter* DataCenter::getInstance()
{
    if (!m_instanceFlag)
    {
        m_instance = new DataCenter();
        m_instanceFlag = true;
    }    
    return m_instance;      
}

我正在进行多线程编程，DataCenter将被超过一个线程。我曾经有方法2获得DataCenter的实例，它工作正常。但我注意到，我需要防止singleton实例被多线程调用。

I am working on a multi threaded programming and DataCenter will be accessed by more than one thread. I used to have method 2 to get the instance of DataCenter and it worked fine. But I noted that I need to guard the singleton instance from being called by multi threads.

我的问题是我真的需要保护单例实例吗？或者OS为我做这个？第二个问题是，第一个方法是正确的方式来获得单例实例吗？

My question is first do I really need to guard the singleton instance? or does OS do this for me? Second questions is that, is the first method a right way to get the singleton instance?

提前感谢...

推荐答案

需要保护它，即使你不，当然，操作系统不会为你做。使用以下代码来确保线程安全：

1.You do need to guard it and even if you don't, of course, OS wouldn't do it for you. Use following code for thread-safety:

DataCenter* DataCenter::getInstance()
{
    MutexLocker locker(DataCenter::m_mutex);
    if(!m_instanceFlag)
    {
        m_instance = new DataCenter();
        m_instanceFlag = true;
    }
    return m_instance;
}

编辑：

其中 MutexLocker 如下所示：

class MutexLocker
{
    pthread_mutex_t &mutex;
    public:
    MutexLocker(pthread_mutex_t &mutex):mutex(mutex)
    {
        if(pthread_mutex_lock(&this->mutex)!=0)
            throw std::runtime_error("mutex locking filed");
    }
    ~MutexLocker(void)
    {
        if(pthread_mutex_unlock(&this->mutex)!=0)
            throw std::runtime_error("mutex unlocking filed");
    }
}

2.第一种方法看起来不错，安全。

2.First method looks ok, but not thread-safe.

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