如何获取自C ++（MSVS）以来的本地当前时间（以秒为单位）？

Question

I need the local (with timezone offset) current time in seconds since epoch. The following code looks a bit clumzy because it creates an intermediate temporary structure tm which is superfluous. Why do I have to get time_t then convert it to tm in order to return to time_t ? Is there a better way?

time_t ct = time(0);
tm lct = tm();
localtime_s(&lct, &ct);
ct = _mkgmtime(&lct);

Answer 1

If you want to get the local time (with time zone and DST applied) in portable C, then yes, it's generally a two-step procedure: starting with your time-since-the-epoch, first call localtime , then do something with the resulting broken-down struct tm . (Usually what I do next is call strftime .)

You can also call ctime to get a local time string directly.

The reason there are a lot of different function calls involved is that, unfortunately, there are several different time formats in use. (And the reason for that is that dates and times are complicated!) You can represent time as seconds-since-1970. You can represent it as a struct tm . You can represent it as a string (in one of several zillion formats). In Unix and Linux, you can represent it as a struct timeval or a struct timespec .

But one thing there isn't a straightforward or standard way to do, as you've discovered, is get local time as seconds-since-1970. But the reason for that is that it's not a very useful representation. In general, there are two things you might want to do with a date/time value: (1) perform computations on it or (2) display it to the user. If you want to display it to the user, you probably want to display it in local time, so there are lots of ways of converting to local time in human-readable format in any format you want. (As I said, the usual way is to call localtime , then strftime .) But if you want to perform computations, really the only way to do those is using seconds-since-1970 in UTC, because that makes all the other hairy problems go away. How many days are there in the month? Is it a leap year? What time zone are we in? Is daylight saving time in effect?

If you try to represent local time as seconds-since-1970, though, you're probably fibbing. For example, right now, the time is 1460383736, which is 14:08:56 UTC. Where I'm sitting, that's 10:08:56 EDT (US Eastern time, DST in effect). So I suppose I could say that's 1460369336 seconds since 1970, local time. But, again where I'm sitting, 1460369336 seconds ago was not midnight on January 1, 1970 -- it was actually 11 pm on December 31, 1969. It's off by an hour, and the reason is that DST was not in effect on January 1, 1970.

So, bottom line, I would encourage you to rethink the way you're handling local times, because while it's possible to compute this "seconds-since-1970 as local time" value, it's an odd thing to do, and it's likely to cause you various problems which will be much harder to work around than if you used a more straightforward scheme.

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But, if you really want to, here are two ways you might be able to determine the offset between UTC and local time, without calling gmtime or _mkgmtime :

1. Call localtime , and look at the tm_gmtoff field. (Unfortunately, this field is nonstandard, and not present on all systems.)
2. Call the obsolete ftime function, and look at the timezone field of struct timeb . (Here there are several gotchas: not only is ftime obsolete and nonstandard, but the timezone field is in minutes , and it's positive for zones west of Greenwich, while tm_gmtoff is negative.)

But, anyway, those would more or less directly give you the number to add to or subtract from your UTC seconds-since-1970 value to get "local" seconds-since-1970.

Answer 2

Here is a way to do this computation using the C++11/14 <chrono> library plus this free, open-source timezone library to do the conversion to local time.

#include "tz.h"
#include <iostream>

int
main()
{
    using namespace date;
    using namespace std;
    using namespace std::chrono;
    auto now = floor<seconds>(system_clock::now());
    auto s = current_zone()->to_local(now) - local_days{1970_y/jan/1};
    cout << s.count() << '\n';
}

You first discover your current IANA timezone with current_zone() . Then you get the current time with system_clock::now() and truncate it to seconds . Next you can convert that to your local time, and then subtract the result from any epoch you desire (1970-01-01 in this example).

The result is of type std::chrono::seconds .

All this being said, I share the same reservations about doing this as described in Steve Summit's answer .

If you instead decide to represent the timestamp as a string, that is also easily done:

auto now = make_zoned(current_zone(), floor<seconds>(system_clock::now()));
auto str = format("%F %T %z", now);

str has type std::string . This just output for me:

2016-04-11 11:42:50 -0400

which is my current local time (truncated to seconds), and my current local UTC offset.

If in the future you decide that seconds-precision is too coarse, you can easily change the above code to any other precision by just changing one line:

floor<milliseconds>(system_clock::now());

and now the contents of str would look like:

2016-04-11 11:42:50.368 -0400