我正在编写一个程序,从用户那里获取输入日期,使用它来初始化tm struct
,然后使用chrono::time_points
执行一些chrono::duration
操作,例如获取年龄。
这是我的代码:
#include <iostream>
#include <chrono>
#include <ctime>
#include <iomanip>
using namespace std;
int main(){
//representing a date
tm *birthday = new tm{00, 30, 00, 07, 11, 97};
//convert to time t
time_t bt = mktime(birthday);
//convert time_t to time_point
chrono::system_clock::time_point t = chrono::system_clock::from_time_t(bt);
chrono::system_clock::time_point now = chrono::system_clock::now();
/*.. Testing
time_t nn = chrono::system_clock::to_time_t(now);
time_t tnn = chrono::system_clock::to_time_t(t);
*/
chrono::system_clock::duration lft = now - t;
//convert to timepoint
chrono::system_clock::time_point tlft{lft};
time_t lifetime = chrono::system_clock::to_time_t(tlft);
cout << put_time(localtime(&lifetime), "%F %T") << endl;
return 0;
}
我的输出是这样的:
$> 1990-02-10 09:42:46
所以,根据我的理解,它在刻度线上执行简单的数学减法并使用localtime,将其转换为EPOCH
之后的日期,这就是它给我1990年的原因。我想知道,是否有任何方式,将持续时间直接转换为struct tm
,以便我得到20年的东西?
答案 0 :(得分:2)
以下是以您选择的单位提取持续时间的方法:
std::chrono::duration<double> lft = now - t;
using year = std::chrono::duration<int, std::ratio<31557600>>;
auto nby = std::chrono::duration_cast<year>(lft);
std::cout << nby.count() << "\n";
考虑到这一点,我建议实施以下的品味:
struct Age
{
using year = std::chrono::duration<int, std::ratio<31'557'600>>;
using month = std::chrono::duration<int, std::ratio< 2'592'000>>;
using day = std::chrono::duration<int, std::ratio< 86'400>>;
using hour = std::chrono::hours;
using minute = std::chrono::minutes;
using second = std::chrono::seconds;
Age(std::chrono::system_clock::time_point birth)
: _age(std::chrono::system_clock::now() - birth)
{}
template<class Duration>
auto extract()
{
const auto result = std::chrono::duration_cast<Duration>(_age);
_age -= result;
return result;
}
friend std::ostream& operator<<(std::ostream& os, Age age)
{
const auto years = age.extract<year>();
const auto monthes = age.extract<month>();
const auto days = age.extract<day>();
const auto hours = age.extract<hour>();
const auto minutes = age.extract<minute>();
const auto seconds = age.extract<second>();
return os << years.count()
<< ":" << std::setw(2) << std::setfill('0') << monthes.count()
<< ":" << std::setw(2) << std::setfill('0') << days.count()
<< " " << std::setw(2) << std::setfill('0') << hours.count()
<< ":" << std::setw(2) << std::setfill('0') << minutes.count()
<< ":" << std::setw(2) << std::setfill('0') << seconds.count()
;
}
private:
std::chrono::duration<double> _age;
};