C++ 将秒作为双精度转换为 std::chrono::duration？
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convert seconds as double to std::chrono::duration?
提问by KaiserJohaan
I'm using c++11 <chrono>and have a number of seconds represented as a double. I want to use c++11 to sleep for this duration, but I cannot fathom how to convert it to a std::chrono::durationobject that std::this_thread::sleep_forrequires.
我正在使用 c++11<chrono>并且有一些秒数表示为双精度数。我想在这段时间内使用 c++11 来睡眠，但我无法理解如何将它转换为需要的std::chrono::duration对象std::this_thread::sleep_for。
const double timeToSleep = GetTimeToSleep();
std::this_thread::sleep_for(std::chrono::seconds(timeToSleep)); // cannot convert from double to seconds
I've locked at the <chrono>reference but I find it rather confusing.
我已经锁定了<chrono>参考文献，但我发现它相当混乱。
Thanks
谢谢
EDIT:
编辑：
The following gives error:
以下给出错误：
std::chrono::duration<double> duration(timeToSleep );
std::this_thread::sleep_for(duration);
the error:
错误：
c:\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
采纳答案by Cornstalks
Don't do std::chrono::seconds(timeToSleep). You want something more like:
不要做std::chrono::seconds(timeToSleep)。你想要更像的东西：
std::chrono::duration<double>(timeToSleep)
Alternatively, if timeToSleepis not measured in seconds, you can pass a ratio as a template parameter to duration. See here(and the examples there) for more information.
或者，如果timeToSleep不是以秒为单位测量，您可以将比率作为模板参数传递给duration。有关更多信息，请参见此处（以及此处的示例）。
回答by Alastair Harrison
Making the answer from @Cornstalks a little more generic, you could define a function like this:
使@Cornstalks 的答案更加通用，您可以定义这样的函数：
template <typename T>
auto seconds_to_duration(T seconds) {
return std::chrono::duration<T, std::ratio<1>>(seconds);
}
This will convert a seconds value of any primitive type to a chrono duration. Use it like this:
这会将任何原始类型的秒值转换为计时持续时间。像这样使用它：
const double timeToSleep = GetTimeToSleep();
std::this_thread_sleep_for(seconds_to_duration(timeToSleep));
回答by slaterade
const unsigned long timeToSleep = static_cast<unsigned long>( GetTimeToSleep() * 1000 );
std::this_thread::sleep_for(std::chrono::milliseconds(timeToSleep));
回答by P0W
std::chrono::milliseconds duration(timeToSleep);
std::this_thread::sleep_for( duration );
