获取系统时间作为一天中的秒数

Question

我正在编写一个程序，将时间戳放在相机拍摄的图像上。为了做到这一点，我使用的是Windows 7系统时间。我在下面的代码中使用了GetSystemTimeAsFileTime():

FILETIME ft;
GetSystemTimeAsFileTime(&ft);
long long ll_now = (LONGLONG)ft.dwLowDateTime + ((LONGLONG)(ft.dwHighDateTime) << 32LL);

我想做的是用毫秒分辨率获得一天中(0- 86400)的秒数，所以它将是12345.678。这是正确的做法吗?如果是这样，我如何转换这个整数，以获得秒数在当前的一天?我将在字符串中显示时间，并使用fstream将时间放在文本文件中。

谢谢

Answer 1

我不知道Window API，但C++标准库(自c++ 11)可以这样使用:

#include <ctime>
#include <chrono>
#include <string>
#include <sstream>
#include <iomanip>
#include <iostream>
std::string stamp_secs_dot_ms()
{
    using namespace std::chrono;
    auto now = system_clock::now();
    // tt stores time in seconds since epoch
    std::time_t tt = system_clock::to_time_t(now);
    // broken time as of now
    std::tm bt = *std::localtime(&tt);
    // alter broken time to the beginning of today
    bt.tm_hour = 0;
    bt.tm_min = 0;
    bt.tm_sec = 0;
    // convert broken time back into std::time_t
    tt = std::mktime(&bt);
    // start of today in system_clock units
    auto start_of_today = system_clock::from_time_t(tt);
    // today's duration in system clock units
    auto length_of_today = now - start_of_today;
    // seconds since start of today
    seconds secs = duration_cast<seconds>(length_of_today); // whole seconds
    // milliseconds since start of today
    milliseconds ms = duration_cast<milliseconds>(length_of_today);
    // subtract the number of seconds from the number of milliseconds
    // to get the current millisecond
    ms -= secs;
    // build output string
    std::ostringstream oss;
    oss.fill('0');
    oss << std::setw(5) << secs.count();
    oss << '.' << std::setw(3) << ms.count();
    return oss.str();
}
int main()
{
    std::cout << stamp_secs_dot_ms() << 'n';
}

示例输出:

13641.509