如何将 std::chrono::time_point 转换为带小数秒的日历

How to convert std::chrono::time_point to calendar datetime string with fractional seconds?(如何将 std::chrono::time_point 转换为带小数秒的日历日期时间字符串?)

问题描述

如何将 std::chrono::time_point 转换为带小数秒的日历日期时间字符串?

How to convert std::chrono::time_point to calendar datetime string with fractional seconds?

例如:

"10-10-2012 12:38:40.123456"

推荐答案

如果是system_clock，这个类有time_t转换.

If system_clock, this class have time_t conversion.

#include <iostream>
#include <chrono>
#include <ctime>

using namespace std::chrono;

int main()
{
  system_clock::time_point p = system_clock::now();

  std::time_t t = system_clock::to_time_t(p);
  std::cout << std::ctime(&t) << std::endl; // for example : Tue Sep 27 14:21:13 2011
}

示例结果:

Thu Oct 11 19:10:24 2012

但是， time_t 不包含小数秒.另一种方法是使用 time_point::time_since_epoch() 函数.此函数返回纪元的持续时间.下面的例子是毫秒分辨率的小数.

But, time_t does not contain fractional seconds. Alternative way is to use time_point::time_since_epoch() function. This function returns duration from epoch. Follow example is milli second resolution's fractional.

#include <iostream>
#include <chrono>
#include <ctime>

using namespace std::chrono;

int main()
{
  high_resolution_clock::time_point p = high_resolution_clock::now();

  milliseconds ms = duration_cast<milliseconds>(p.time_since_epoch());

  seconds s = duration_cast<seconds>(ms);
  std::time_t t = s.count();
  std::size_t fractional_seconds = ms.count() % 1000;

  std::cout << std::ctime(&t) << std::endl;
  std::cout << fractional_seconds << std::endl;
}

示例结果:

Thu Oct 11 19:10:24 2012

925

这篇关于如何将 std::chrono::time_point 转换为带小数秒的日历日期时间字符串?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持编程学习网！