問題描述

好吧，我在使用 std::thread 將數據傳遞到線程時遇到了問題.我以為我了解復制構造函數等的一般語義，但似乎我不太明白這個問題.我有一個名為 Log 的簡單類，它因此隱藏了它的復制構造函數:

Well I have an issue with passing data into a thread using std::thread. I thought I understood the general semantics of copy constructors, etc. but it seems I don't quite grasp the problem. I have a simple class called Log that has hidden it's copy constructor thusly:

class Log
{
public:
    Log(const char filename[], const bool outputToConsole = false);
    virtual ~Log(void);

    //modify behavior
    void appendStream(std::ostream *);
    //commit a new message
    void commitStatus(const std::string str);

private:
    //members
    std::ofstream fileStream;
    std::list<std::ostream *> listOfStreams;

    //disable copy constructor and assignment operator
    Log(const Log &);
    Log & operator=(const Log &);
}

現在我有一個主要基于 http://www.boost.org/doc/libs/1_55_0/doc/html/boost_asio/example/cpp11/echo/blocking_tcp_echo_server.cpp

now I have a main based heavily on http://www.boost.org/doc/libs/1_55_0/doc/html/boost_asio/example/cpp11/echo/blocking_tcp_echo_server.cpp

int main()
{
    static int portNumber = 10000;

    Log logger("ServerLog.txt", true);
    logger.commitStatus("Log Test String");

    try {
        boost::asio::io_service ioService;
        server(ioService, portNumber, logger);
    }
    catch (std::exception &e)
    {
        std::cerr << "Exception " << e.what() << std::endl;
        logger.commitStatus(e.what());
    }

    return 0;
}

可以看到main調用了函數server，傳遞了IOService、portNumber和logger.記錄器是通過引用傳遞的，因此:

You can see that main calls the function server and passes the IOService, portNumber and logger. The logger is passed by reference, thusly:

using boost::asio::ip::tcp;

void server(boost::asio::io_service &ioService, unsigned int port, Log &logger)
{
    logger.commitStatus("Server Start");

    tcp::acceptor acc(ioService, tcp::endpoint(tcp::v4(), port));

    while(true)
    {
        tcp::socket sock(ioService);
        acc.accept(sock);

        std::thread newThread(session, &sock, logger);
        newThread.detach();
    }

    logger.commitStatus("Server closed");
}

當我嘗試通過引用將記錄器(或套接字)傳遞給線程時出現編譯器錯誤，但通過引用將它傳遞給 session() 時我沒有收到錯誤

I get a compiler error when I try to pass the logger (or the socket) to the thread by reference, but I do not get the error when passing it to the session() by reference

static void session(tcp::socket *sock, Log &logger)
{
    std::cout << " session () " << std::endl;
}

現在我認為我正確理解了引用與傳遞指針相同.也就是說，它不調用復制構造函數，它只是傳遞指針，它讓您在語法上將其視為不是指針.

Now I thought that I understood correctly that a reference is the same as passing a pointer. That is, it does not call the copy constructor, it simply passes the pointer, which it lets you syntactically treat like it's not a pointer.

錯誤 C2248:'Log::Log':無法訪問類 'Log' 中聲明的私有成員

error C2248: 'Log::Log' : cannot access private member declared in class 'Log'

1> log.h(55) : 見 'Log::Log' 的聲明

1> log.h(55) : see declaration of 'Log::Log'

1> log.h(28) : 見 'Log' 的聲明

1> log.h(28) : see declaration of 'Log'

...

:參見正在編譯的函數模板實例化'std::thread::thread(_Fn,_V0_t &&,_V1_t)'的參考

: see reference to function template instantiation 'std::thread::thread(_Fn,_V0_t &&,_V1_t)' being compiled

1> 與

1> [

1> Fn=void (_cdecl *)(boost::asio::ip::tcp::socket *,Log &),

1> Fn=void (_cdecl *)(boost::asio::ip::tcp::socket *,Log &),

1> _V0_t=boost::asio::ip::tcp::socket *,

1> _V0_t=boost::asio::ip::tcp::socket *,

1> _V1_t=日志 &

1> _V1_t=Log &

1>]

但是如果我修改它來傳遞一個指針，一切都會很開心

However if I modify it to pass a pointer, everything is happy

...
        std::thread newThread(session, &sock, &logger);
...

static void session(tcp::socket *sock, Log *logger)
{
    std::cout << " session () " << std::endl;
}

為什么通過引用傳遞調用我的復制構造函數.由于std::thread，這里有什么特別的事情發生嗎?我是否誤解了復制構造函數并通過引用傳遞?

Why is passing by reference calling my copy constructor. Is there something special happening here because of std::thread? Did I misunderstand the copy constructor and pass by reference?

如果我像示例中那樣嘗試使用 std::move() ，我會得到一個不同但同樣令人困惑的錯誤.我的 VS2012 是否可能沒有正確實現 C++11?

I get a different but equally baffling error if I try to use std::move() as it is done in the example. Is it possible my VS2012 is not implementing C++11 correctly?

推薦答案

std::thread 按值接受其參數.你可以通過使用 std::reference_wrapper:

std::thread takes its arguments by value. You can get reference semantics back by using std::reference_wrapper:

std::thread newThread(session, &sock, std::ref(logger));

顯然，您必須確保 logger 比線程壽命更長.

Obviously you must make sure that logger outlives the thread.

這篇關于std::thread 通過引用調用復制構造函數的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！