How to print time in format: 2009‐08‐10 18:17:54.811

trick:

    int time_len = 0, n;
    struct tm *tm_info;
    struct timeval tv;

    gettimeofday(&tv, NULL);
    tm_info = localtime(&tv.tv_sec);
    time_len+=strftime(log_buff, sizeof log_buff, "%y%m%d %H:%M:%S", tm_info);
    time_len+=snprintf(log_buff+time_len,sizeof log_buff-time_len,".%03ld ",tv.tv_usec/1000);

None of the solutions on this page worked for me, I mixed them up and made them working with Windows and Visual Studio 2019, Here's How :

#include <Windows.h>
#include <time.h> 
#include <chrono>

static int gettimeofday(struct timeval* tp, struct timezone* tzp) {
    namespace sc = std::chrono;
    sc::system_clock::duration d = sc::system_clock::now().time_since_epoch();
    sc::seconds s = sc::duration_cast<sc::seconds>(d);
    tp->tv_sec = s.count();
    tp->tv_usec = sc::duration_cast<sc::microseconds>(d - s).count();
    return 0;
}

static char* getFormattedTime() {
    static char buffer[26];

    // For Miliseconds
    int millisec;
    struct tm* tm_info;
    struct timeval tv;

    // For Time
    time_t rawtime;
    struct tm* timeinfo;

    gettimeofday(&tv, NULL);

    millisec = lrint(tv.tv_usec / 1000.0);
    if (millisec >= 1000) 
    {
        millisec -= 1000;
        tv.tv_sec++;
    }

    time(&rawtime);
    timeinfo = localtime(&rawtime);

    strftime(buffer, 26, "%Y:%m:%d %H:%M:%S", timeinfo);
    sprintf_s(buffer, 26, "%s.%03d", buffer, millisec);

    return buffer;
}

Result :

2020:08:02 06:41:59.107

2020:08:02 06:41:59.196