多个线程同时在 unordered_map 中添加值使其崩溃

Question

unordered_map<std::string,unordered_map<std::string, std::string> >* storing_vars;

我在作用域中声明了这个变量。

这是在构造函数中声明的。

this->storing_vars =  new unordered_map<std::string,unordered_map<std::string, std::string> >();

为了初始化它。

然后我所做的就是由我的 BackgroundWorker 一遍又一遍地调用一个函数

for(int i2 = 0; i2 < 30; i2++){



                int index_pos_curr = i2;


                //Start the Threads HERE


                this->backgroundWorker2 = gcnew System::ComponentModel::BackgroundWorker;
                this->backgroundWorker2->WorkerReportsProgress = true;
                this->backgroundWorker2->WorkerSupportsCancellation = true;


                //this->backgroundWorker2->FieldSetter(L"std::string",L"test","damnnit");

                backgroundWorker2->DoWork += gcnew DoWorkEventHandler( this, &MainFacebook::backgroundWorker2_DoWork );
                backgroundWorker2->RunWorkerCompleted += gcnew RunWorkerCompletedEventHandler( this, &MainFacebook::backgroundWorker2_RunWorkerCompleted );
                backgroundWorker2->ProgressChanged += gcnew ProgressChangedEventHandler( this, &MainFacebook::backgroundWorker2_ProgressChanged );
                backgroundWorker2->RunWorkerAsync(index_pos_curr);
                Sleep(50); //THE PROBLEM IS HERE, IF I COMMENT THIS OUT it won't work, that's probably because there are a lot of functions trying to add values in the same variable (even though the indexes are differents in each call)
            }

完成此操作后，它会调用 DoWork 函数

void backgroundWorker2_DoWork(Object^ sender, DoWorkEventArgs^ e ){
             BackgroundWorker^ worker = dynamic_cast<BackgroundWorker^>(sender);         
             e->Result = SendThem( safe_cast<Int32>(e->Argument), worker, e );          

        }

int SendThem(int index){

            stringstream st;
            st << index;

            //...
            (*this->storing_vars)[st.str()]["index"] =  "testing1";
            (*this->storing_vars)[st.str()]["rs"] = "testing2";
            return 0;
}

当我在Sleep(50) 行中添加注释时，我认为问题在于，由于后台线程调用相同的函数，因此在多次调用时存储数据存在问题，甚至可能没有等待其他存储完成，这会导致“xhash.h”文件中出现错误，该错误已通过使用Sleep(50) 进行清理，但我无法使用这些，因为它会冻结我的 UI，而且 50 毫秒是时间我假设它已经存储了变量值，但是如果在较慢的计算机上需要更长的时间呢？这不是正确的方法。

我该如何解决这个问题？

我希望能够在不使用 SLEEP 的情况下更新 unordered_map

提前致谢。

Answer 1

您一次只能从一个线程修改标准库容器（包括但不限于unordered_map）。解决方案是使用临界区、互斥体、锁来同步访问。如果您不知道这些是什么，那么您需要在尝试创建多个线程之前了解。

没有如果、但是或为什么。

如果您有多个线程，则需要同步它们的机制，以序列化对共享数据的访问。常见的同步机制就是上面提到的那些，你去看看吧。

Answer 2

经过这么多票否决后，我实际上开始寻找互斥锁，人们在这里谈论，过了一段时间我发现它真的很容易使用。这是我的同事告诉我的正确方法。谢谢大家的帮助=D

这是我所做的，我只需要添加

//Declare the Mutex
static Mutex^ mut = gcnew Mutex;

//then inside the function called over and over again I used mutex
mut->WaitOne();
//Declare/Update the variables
mut->ReleaseMutex();
//Then I release it.

效果很好，谢谢大家的帮助和批评。哈哈

Answer 3

我通过预定义我想使用的 unordered_map 的索引找到了一种解决方案，问题只是创建索引，多线程更新似乎没问题。

for(int i2 = 0; i2 < 30; i2++){



                int index_pos_curr = i2;


                //Start the Threads HERE


                this->backgroundWorker2 = gcnew System::ComponentModel::BackgroundWorker;
                this->backgroundWorker2->WorkerReportsProgress = true;
                this->backgroundWorker2->WorkerSupportsCancellation = true;
                backgroundWorker2->DoWork += gcnew DoWorkEventHandler( this, &MainFacebook::backgroundWorker2_DoWork );
                backgroundWorker2->RunWorkerCompleted += gcnew RunWorkerCompletedEventHandler( this, &MainFacebook::backgroundWorker2_RunWorkerCompleted ); stringstream st; st << index_pos_curr;

                (*this->storing_vars)[st.str()]["index"] = ""; 
               //This ^^^^^ will initialize it and then in the BackgroundWorker will only update, this way it doesn't crash. :)

                backgroundWorker2->ProgressChanged += gcnew ProgressChangedEventHandler( this, &MainFacebook::backgroundWorker2_ProgressChanged );
                backgroundWorker2->RunWorkerAsync(index_pos_curr);
                Sleep(50); //THE PROBLEM IS HERE, IF I COMMENT THIS OUT it won't work, that's probably because there are a lot of functions trying to add values in the same variable (even though the indexes are differents in each call)
            }