简单的C++线程

我试图用C++(Win32)创建一个线程来运行一个简单的方法.我是C++线程的新手,但对C#中的线程非常熟悉.这是我正在尝试做的一些伪代码:

static void MyMethod(int data)
{
    RunStuff(data);
}

void RunStuff(int data)
{
    //long running operation here
}

我想从MyMethod调用RunStuff而不阻塞它.在单独的线程上运行RunStuff最简单的方法是什么？

编辑:我还应该提一下,我希望将依赖关系保持在最低限度.(没有MFC ......等)

#include 

static boost::thread runStuffThread;

static void MyMethod(int data)
{
    runStuffThread = boost::thread(boost::bind(RunStuff, data));
}

// elsewhere...
runStuffThread.join(); //blocks

C++ 11可与更新的编译器(如Visual Studio 2013)一起使用,其中包含线程作为语言的一部分以及其他一些非常好的部分,例如lambda.

include文件threads提供了一个线程类,它是一组模板.线程功能在std::命名空间中.一些线程同步函数std::this_thread用作命名空间(请参阅为什么std :: this_thread命名空间？以获得一些解释).

使用Visual Studio 2013的以下控制台应用程序示例演示了C++ 11的一些线程功能,包括使用lambda(请参阅C++ 11中的什么是lambda表达式？).请注意,用于线程休眠的函数(例如std::this_thread::sleep_for(),使用持续时间来自)std::chrono.

// threading.cpp : Defines the entry point for the console application.
//

#include "stdafx.h"
#include 
#include 
#include 
#include 

int funThread(const char *pName, const int nTimes, std::mutex *myMutex)
{
    // loop the specified number of times each time waiting a second.
    // we are using this mutex, which is shared by the threads to
    // synchronize and allow only one thread at a time to to output.
    for (int i = 0; i < nTimes; i++) {
        myMutex->lock();
        std::cout << "thread " << pName << " i = " << i << std::endl;
        // delay this thread that is running for a second.  
        // the this_thread construct allows us access to several different
        // functions such as sleep_for() and yield().  we do the sleep
        // before doing the unlock() to demo how the lock/unlock works.
        std::this_thread::sleep_for(std::chrono::seconds(1));
        myMutex->unlock();
        std::this_thread::yield();
    }

    return 0;
}

int _tmain(int argc, _TCHAR* argv[])
{
    // create a mutex which we are going to use to synchronize output
    // between the two threads.
    std::mutex  myMutex;

    // create and start two threads each with a different name and a
    // different number of iterations. we provide the mutex we are using
    // to synchronize the two threads.
    std::thread myThread1(funThread, "one", 5, &myMutex);
    std::thread myThread2(funThread, "two", 15, &myMutex);

    // wait for our two threads to finish.
    myThread1.join();
    myThread2.join();

    auto  fun = [](int x) {for (int i = 0; i < x; i++) { std::cout << "lambda thread " << i << std::endl; std::this_thread::sleep_for(std::chrono::seconds(1)); } };

    // create a thread from the lambda above requesting three iterations.
    std::thread  xThread(fun, 3);
    xThread.join();
    return 0;
}