C++ 在类中使用静态互斥锁
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Using static mutex in a class
提问by Dmitry Yudakov
I have a class that I can have many instances of. Inside it creates and initializes some members from a 3rd party library (that use some global variables) and is not thread-safe.
我有一个类，我可以有很多实例。在它内部创建和初始化来自 3rd 方库（使用一些全局变量）的一些成员并且不是线程安全的。
I thought about using static boost::mutex, that would be locked in my class constructor and destructor. Thus creating and destroying instances among my threads would be safe for the 3rd party members.
我想过使用静态 boost::mutex，它会被锁定在我的类构造函数和析构函数中。因此，在我的线程中创建和销毁实例对 3rd 方成员来说是安全的。
class MyClass
{
static boost::mutex mx;
// 3rd party library members
public:
MyClass();
~MyClass();
};
MyClass::MyClass()
{
boost::mutex::scoped_lock scoped_lock(mx);
// create and init 3rd party library stuff
}
MyClass::~MyClass()
{
boost::mutex::scoped_lock scoped_lock(mx);
// destroy 3rd party library stuff
}
I cannot link because I receive error:
我无法链接，因为我收到错误：
undefined reference to `MyClass::mx`
Do I need some special initialization of such static member?
Is there anything wrong about using static mutex?
我需要对这种静态成员进行一些特殊的初始化吗？
使用静态互斥量有什么问题吗？
Edit:Linking problem is fixed with correct definition in cpp
编辑：链接问题已通过 cpp 中的正确定义解决
boost::mutex MyClass::mx;
回答by Stephen C. Steel
You have declared, but not defined your class static mutex. Just add the line
您已声明但未定义您的类静态互斥锁。只需添加行
boost::mutex MyClass::mx;
to the cpp file with the implementation of MyClass.
到带有 MyClass 实现的 cpp 文件。
