C++ Meyers Singleton - 线程安全(互斥量的等效代码?) [英] C++ Meyers Singleton - thread safe (code equivalent for mutex?)

问题描述

上周有人向我指出了一段这样的代码:

I was pointed last week about a piece of code like this:

#include <pthread.h>
namespace NSTest
{
class SingletonClass
{
public:


    static SingletonClass & getInstance()
    {
        static  pthread_mutex_t mutex;
        pthread_mutex_lock(&mutex);
        if(singletonPtr==nullptr)
        {
            createInstence();
        }
        return (*singletonPtr);
        pthread_mutex_unlock(&mutex);

    }

private:
    static void createInstence()
    {
        static SingletonClass static_SingletonClass;
        singletonPtr=&static_SingletonClass;

    }
    static SingletonClass * singletonPtr;

};

SingletonClass * SingletonClass::singletonPtr=nullptr;


class SingletonClassNoStatic
{
public:
    static SingletonClassNoStatic & getInstance()
    {
        pthread_mutex_lock(&mutex);
        if(singletonPtr==nullptr)
        {
            createInstence();
        }
        return (*singletonPtr);
        pthread_mutex_unlock(&mutex);
    }
private:
    static void createInstence()
    {
        static SingletonClassNoStatic static_SingletonClass;
        singletonPtr=&static_SingletonClass;

    }
    static SingletonClassNoStatic * singletonPtr;

    static  pthread_mutex_t mutex;
};
SingletonClassNoStatic * SingletonClassNoStatic::singletonPtr=nullptr;
pthread_mutex_t SingletonClassNoStatic::mutex;
}
int main()
{

    NSTest::SingletonClass::getInstance();
    NSTest::SingletonClassNoStatic::getInstance();

    return 0;
}

方法 getInstance ，被指出是正确的，原来是 getIntance(StaticMutex)，编码不是线程安全的 (c++98) 因为互斥量是在方法中创建的，它是静态的，并且使用静态互斥锁，我遵循静态方法中的静态实例的规则创建一次，不再创建，但此规则自不应用于互斥锁.我有疑问，改正可以吗?每个访问该方法(静态互斥锁)的线程都会创建自己的互斥锁?我读到该操作仅适用于类中的方法，但是因为互斥锁是在静态方法中静态创建的，所以它将被创建一次.我是否正确理解了这个概念?

The method getInstance , was pointed to be the correct one, and the original was getIntance(StaticMutex), that coded is not thread safe (c++98) because the mutex is created in the method , that it was static and with a static mutex, I was following the rule of the static instance in static method is created once and no longer is created, but this rule since not to be applied to the mutex. I have doubts, it that correction be ok? Each thread that accesses that method (static mutex) will create its own mutex? I was reading that that action is only to method that are method from class, but because the mutex is created static in a static method it will be created once. Am I correctly understanding the concept?

推荐答案

在 C++11(由 std::mutex 隐含)中，整个 Meyers 单例用两行表示:

In C++11 (which is implied by std::mutex) the whole Meyers singleton is expressed in two lines:

Singleton& instance() {
  static Singleton instance;

  return instance;
}

不需要其他任何东西.问题本身很不清楚，因为您在同一上下文中谈论 C++98 和 std::mutex，而这些并没有加起来.

Nothing else is needed. The question itself is very unclear, since you talk about C++98 and std::mutex in the same context, and those do not add up.

发布代码的另一个问题是它不是开始是迈耶的单例.

The other problem with posted code is that it is not a Meyer's singleton to begin with.

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