我正在编写一个C代码,需要访问一个使用timeval作为当前时间表示的旧C库.
在旧包中获取我们使用的当前日期/时间：
struct timeval dateTime; gettimeofday(&dateTime,NULL); function(dateTime); // The function will do its task
现在我需要使用C计时器,如：
system_clock::time_point now = system_clock::now();
struct timeval dateTime;
dateTime.tv_sec = ???? // Help appreaciated here
dateTime.tv_usec = ???? // Help appreaciated here
function(dateTime);
后来在代码中我需要返回的方式,从返回的struct timeval构建一个time_point变量：
struct timeval dateTime;
function(&dateTime);
system_clock::time_point returnedDateTime = ?? // Help appreacited
我使用C11.
解决方法
[编辑为使用time_val而不是free vars]
假设你以毫秒的精度信任你的system_clock,你可以这样去：
struct timeval dest;
auto now=std::chrono::system_clock::now();
auto millisecs=
std::chrono::duration_cast<std::chrono::milliseconds>(
now.time_since_epoch()
);;
dest.tv_sec=millisecs.count()/1000;
dest.tv_usec=(millisecs.count()%1000)*1000;
std::cout << "s:" << dest.tv_sec << " usec:" << dest.tv_usec << std::endl;
在duration_cast中使用std :: chrono :: microseconds,并相应地调整您的(div / mod)代码以获得更高的精度 – 注意您信任您获得的值的准确性.
转换回来是：
timeval src;
// again,trusting the value with only milliseconds accuracy
using dest_timepoint_type=std::chrono::time_point<
std::chrono::system_clock,std::chrono::milliseconds
>;
dest_timepoint_type converted{
std::chrono::milliseconds{
src.tv_sec*1000+src.tv_usec/1000
}
};
// this is to make sure the converted timepoint is indistinguishable by one
// issued by the system_clock
std::chrono::system_clock::time_point recovered =
std::chrono::time_point_cast<std::chrono::system_clock::duration>(converted)
;
