std::thread 通过引用调用复制构造函数
好吧,我在使用 std::thread 将数据传递到线程时遇到了问题.我以为我了解复制构造函数等的一般语义,但似乎我不太明白这个问题.我有一个名为 Log 的简单类,它因此隐藏了它的复制构造函数:
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 &);
}
现在我有一个主要基于 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调用了函数server,传递了IOService、portNumber和logger.记录器是通过引用传递的,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");
}
当我尝试通过引用将记录器(或套接字)传递给线程时出现编译器错误,但通过引用将它传递给 session() 时我没有收到错误
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;
}
现在我认为我正确理解了引用与传递指针相同.也就是说,它不调用复制构造函数,它只是传递指针,它让您在语法上将其视为不是指针.
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.
错误 C2248:'Log::Log':无法访问类 'Log' 中声明的私有成员
error C2248: 'Log::Log' : cannot access private member declared in class 'Log'
1> log.h(55) : 见 'Log::Log' 的声明
1> log.h(55) : see declaration of 'Log::Log'
1> log.h(28) : 见 'Log' 的声明
1> log.h(28) : see declaration of 'Log'
...
:参见正在编译的函数模板实例化'std::thread::thread(_Fn,_V0_t &&,_V1_t)'的参考
: see reference to function template instantiation 'std::thread::thread(_Fn,_V0_t &&,_V1_t)' being compiled
1> 与
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>]
但是如果我修改它来传递一个指针,一切都会很开心
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;
}
为什么通过引用传递调用我的复制构造函数.由于std::thread,这里有什么特别的事情发生吗?我是否误解了复制构造函数并通过引用传递?
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?
如果我像示例中那样尝试使用 std::move() ,我会得到一个不同但同样令人困惑的错误.我的 VS2012 是否可能没有正确实现 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?
推荐答案
std::thread
按值接受其参数.你可以通过使用 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));
显然,您必须确保 logger
比线程寿命更长.
Obviously you must make sure that logger
outlives the thread.
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