How decode windows 64bit hex value : little endian time?

Question

I want to decode windows 64bit hex value: little endian time in C++.

I have a value(=FB7E5D6A355DD201) but I don't know how decode this value in C++.

Please tell me a sample code or related site.

Answer 1

In unix systems date/time is often (but not always) recorded as a 32-bit integer indicating seconds past since Epoch (00:00:00, Jan 1, 1970). But for Windows 64-bit LE format, it means "x100 nanoseconds past since 00:00:00.000,000,0 Jan 1, 1601 ". Your hexadecimal number is a raw binary integer, so it can be easily interpreted by scanf .
Here's a piece of stuff that I found from one of my oldest C programs. You can optimize it and add time zone support by yourself.

#include <stdio.h>
int mDays[12]={31,28,31,30,31,30,31,31,30,31,30,31};
const int daySec = 86400;
int yearDays(int y){
    return 365 + (y%400==0 || (y%4==0 && y%100!=0));
    // Pre-Caesar calendar doesn't work well, don't add it
}
void win_u64tod(unsigned long long utime){
    int y=1601, d=0, mo=0, h=0, m=0, s=0, ms=0;
    ms = utime % 10000000;
    utime /= 10000000;

    while (utime >= daySec*yearDays(y)){
        utime -= daySec*yearDays(y);
        y ++;
    }
    if (yearDays(y) == 366) mDays[1]=29;
    d = utime / daySec;
    utime %= daySec;
    while (d >= mDays[mo]){
        d -= mDays[mo];
        mo ++;
    }
    h = utime/3600;
    utime %= 3600;
    m = utime/60;
    s = utime%60;
    printf("%4u-%2u-%2u %2u:%02u:%02u.%07u\n", y, 1+mo, 1+d, h, m, s, ms);
}

int main(){
    union{
        char c[8];
        unsigned long long ull;
    } a;
    int i;
    for (i = 0; i < 8; i ++){
        scanf("%2x", a.c + i);
    }
    win_u64tod(a.ull);
}

Sample input: FB7E5D6A355DD201
Sample output: 2016-12-23 15:58:34.6327803 (Please note time zone is not added)

Reference: DATECONVERT