我正在使用C++ 11 <chrono>
並將秒數表示爲double。我想在這段時間內使用C++ 11進行睡眠,但是我無法理解如何將其轉換爲std::this_thread::sleep_for
需要的std::chrono::duration
對象。將秒轉換爲雙精度到std :: chrono :: duration?
const double timeToSleep = GetTimeToSleep();
std::this_thread::sleep_for(std::chrono::seconds(timeToSleep)); // cannot convert from double to seconds
我鎖定在<chrono>
參考,但我覺得它相當混亂。
感謝
編輯:
下面給出了錯誤:
std::chrono::duration<double> duration(timeToSleep);
std::this_thread::sleep_for(duration);
錯誤:
:\program files (x86)\microsoft visual studio 11.0\vc\include\chrono(749): error C2679: binary '+=' : no operator found which takes a right-hand operand of type 'const std::chrono::duration<double,std::ratio<0x01,0x01>>' (or there is no acceptable conversion)
2> c:\program files (x86)\microsoft visual studio 11.0\vc\include\chrono(166): could be 'std::chrono::duration<__int64,std::nano> &std::chrono::duration<__int64,std::nano>::operator +=(const std::chrono::duration<__int64,std::nano> &)'
2> while trying to match the argument list '(std::chrono::nanoseconds, const std::chrono::duration<double,std::ratio<0x01,0x01>>)'
2> c:\program files (x86)\microsoft visual studio 11.0\vc\include\thread(164) : see reference to function template instantiation 'xtime std::_To_xtime<double,std::ratio<0x01,0x01>>(const std::chrono::duration<double,std::ratio<0x01,0x01>> &)' being compiled
2> c:\users\johan\desktop\svn\jonsengine\jonsengine\src\window\glfw\glfwwindow.cpp(73) : see reference to function template instantiation 'void std::this_thread::sleep_for<double,std::ratio<0x01,0x01>>(const std::chrono::duration<double,std::ratio<0x01,0x01>> &)' being compiled
Cornstalks的答案是正確的。這看起來像VS11中的一個錯誤'std :: this_thread :: sleep_for'。你可以嘗試'std :: this_thread :: sleep_for(std :: chrono:duration_cast(duration))'來解決這個錯誤。我任意選擇毫秒。使用任何作品,但讓''提供轉換而不是滾動自己的作品。 –