chrono

问题 In the example here: https://en.cppreference.com/w/cpp/chrono/high_resolution_clock/now They declared the clock time point with auto . auto start = std::chrono::high_resolution_clock::now(); and the doc says it returns 'A time point representing the current time.' But I am not sure how do declare in my code below because I am used to declaring the variables at the beginning of the function and I don't know what to declare it as. Code has been simplified here to show what I mean. What do I put

问题 I am parsing dates and times in Rust using the chrono crate. The dates and times are from a website in which the date and time are from different sections of the page. The date is shown in the format %d/%m/%Y (example: 27/08/2018). The time is shown with only the hour (example: 12, 10, 21, etc.) I want to store these datetimes as UTC so that I can compute time remaining until a given datetime from now in a "timezone agnostic" way. I know which timezone these datetimes are from (Paris time). I

问题 I would like to use std::chrono in order to find calculate the date in the future based on an expiration period. The expiration period is an integer that specifies "days from now". So how do I use chrono lib in order to find the date after 100 days? 回答1: Say you have a time_point. To add days to that object you can just use operator+ with std::chrono::hours : #include <chrono> std::chrono::system_clock::time_point t = std::chrono::system_clock::now(); std::chrono::system_clock::time_point new

问题 I would like to use std::chrono in order to find calculate the date in the future based on an expiration period. The expiration period is an integer that specifies "days from now". So how do I use chrono lib in order to find the date after 100 days? 回答1: Say you have a time_point. To add days to that object you can just use operator+ with std::chrono::hours : #include <chrono> std::chrono::system_clock::time_point t = std::chrono::system_clock::now(); std::chrono::system_clock::time_point new

问题 I wrote a solution for windows using MSVC2015 where the follow code converts the std::filesystem::last_write_time result time_t: time_t ftime = std::file_time_type::clock::to_time_t(fs::last_write_time("/Path/filename")) It works well. Then when I tried to port the solution to Linux using gcc 9.3 (-std=C++2a) I've got the follow error: Error: 'to_time_t' is not member of 'std::chrono::time_point::clock' {aka 'std::filesystem::__file_clock'} I searched for a solution, but what I found is based

问题 Let's have using duration = std::chrono::steady_clock::duration . I would like to convert duration to double in seconds with maximal precition elegantly. I have found the reverse way (convert seconds as double to std::chrono::duration? ), but it didn't help me finding it out. Alternatively expressed, I want optimally some std function double F(duration) , which returns seconds. Thank you. 回答1: Simply do: std::chrono::duration<double>(d).count() Or, as a function: template <class Rep, class

问题 cppreference.com says The default constructor is defaulted . I also checked the C++14 draft , it said nothing on the default constructor, except the declaration: constexpr duration() = default; When I run the following code, I was surprised. chrono::seconds s; cout << s.count() << endl; Each time I run it, the program prints some arbitrary numbers: 140737364037104 , 140737078676496 and so on. It seems that s does NOT well initialized. Then I checked my compiler (GCC 4.8)'s implementation for

问题 cppreference.com says The default constructor is defaulted . I also checked the C++14 draft , it said nothing on the default constructor, except the declaration: constexpr duration() = default; When I run the following code, I was surprised. chrono::seconds s; cout << s.count() << endl; Each time I run it, the program prints some arbitrary numbers: 140737364037104 , 140737078676496 and so on. It seems that s does NOT well initialized. Then I checked my compiler (GCC 4.8)'s implementation for

问题 If I convert to a coarser unit of time (say std::chrono::minutes to std::chrono::hours ) how will duration_cast round? For example, what value will std::chrono::minutes(91) become if converted to std::chrono::hours ? 2h, 1h? 回答1: duration_cast always rounds towards zero. I.e. positive values round down and negative values round up. For other rounding options see: http://howardhinnant.github.io/duration_io/chrono_util.html floor , ceil , and round are currently in the draft C++ 1z (hopefully C

问题 I have a template class, something like: template < typename T, size_t Seconds > class MyClass {} Now, I would like to change Seconds to be a duration, so the class can be parametrized with std::chrono::duration . For example, I'd like to be able to do this: MyClass < std::string, std::chrono::seconds(30) > object; Also, in the template, I'd like to specify a default value, something like std::chrono::seconds(30) . 回答1: You can design your template in a clever way: template < typename T,