我有一个以下时间格式的字符串：
“％Y-％m-％d％H：％M：％S.％f”
其中％f是毫秒,例如：14：31：23.946571
我希望这是一个计时时间点.有演员这样做吗？
解决方法
没有从std :: string转换为std :: chrono :: time_point.您必须构建std :: chrono :: time_point对象.
>使用除微秒之外的所有内容来构造std :: tm对象(< ctime>).
年份应该基于1900而不是0.月份应该基于0,而不是1.
>使用std :: mktime()创建std :: time_t对象.
>使用from_time_t()创建一个std :: chrono :: time_point.
>将剩余的小数部分(视为int)添加为time_point的std :: chrono :: microsecond()持续时间.
请注意< iomanip>函数std :: ctime()和std :: put_time()不知道精度低于一秒.如果要打印该精度级别,则需要编写一个函数来执行此操作.
#include <chrono>
#include <ctime>
#include <iomanip>
#include <iostream>
struct Tm : std::tm {
int tm_usecs; // [0,999999] micros after the sec
Tm(const int year,const int month,const int mday,const int hour,const int min,const int sec,const int usecs,const int isDST = -1)
: tm_usecs{usecs} {
tm_year = year - 1900; // [0,60] since 1900
tm_mon = month - 1; // [0,11] since Jan
tm_mday = mday; // [1,31]
tm_hour = hour; // [0,23] since midnight
tm_min = min; // [0,59] after the hour
tm_sec = sec; // [0,60] after the min
// allows for 1 positive leap second
tm_isdst = isDST; // [-1...] -1 for unknown,0 for not DST,// any positive value if DST.
}
template <typename Clock_t = std::chrono::high_resolution_clock,typename MicroSecond_t = std::chrono::microseconds>
auto to_time_point() -> typename Clock_t::time_point {
auto time_c = mktime(this);
return Clock_t::from_time_t(time_c) + MicroSecond_t{tm_usecs};
}
};
int main() {
using namespace std::chrono;
auto tp_nomicro = Tm(2014,8,19,14,31,23,0).to_time_point();
auto tp_micro = Tm(2014,946571).to_time_point();
std::cout << duration_cast<microseconds>(tp_micro - tp_nomicro).count()
<< " microseconds apart.\n";
auto time_c = high_resolution_clock::to_time_t(tp_micro);
std::cout << std::ctime(&time_c) << '\n';
}
