如何获得与C++11计时器完全相同的timeval结果

struct timeval {
    long tv_sec;    // 秒
    long tv_usec;   // 微秒
};

#include <iostream>
#include <chrono>

struct timeval {
    long tv_sec;
    long tv_usec;
};

timeval get_timeval() {
    auto now = std::chrono::system_clock::now();
    auto duration = now.time_since_epoch();
    auto seconds = std::chrono::duration_cast<std::chrono::seconds>(duration);
    auto microseconds = std::chrono::duration_cast<std::chrono::microseconds>(duration - seconds);

    timeval tv;
    tv.tv_sec = seconds.count();
    tv.tv_usec = microseconds.count();
    return tv;
}

int main() {
    timeval tv = get_timeval();
    std::cout << "Seconds: " << tv.tv_sec << ", Microseconds: " << tv.tv_usec << std::endl;
    return 0;
}