Scripting Helpers is winding down operations and is now read-only. More info→

os.time - How to use it?

Asked by

lightpower26 399

10 years ago

I was looking in the blog and it started talking about os.time. I wondered how it outputted so I put in the command line in the in-game console and this happened: 03EFF218 So I tried it in studio and: 0x15af5670 Is this a table or something? Obviously it's not supposed to be a string. How would I use os.time?

Pay close attention to the output: "function: 0D767DE0" the word before the numbers tells you that it's a function (not a table or something) BlueTaslem 18071 — 10y

2 answers

Answered by

BlueTaslem 18071

10 years ago

Time is a very difficult to rigorously define.

The standard way that it is done is by counting the number of seconds from some particular, special moment in history.

tick() counts the number of seconds since the beginning of 1970 in your time zone.

os.clock() does a similar thing. However, it measures universal time meaning it references the beginning of 1970 GMT, regardless of which timezone the computer is in.

os.clock() is also an integer, while tick() provides a fraction of a second.

Here's a function that computes the day, year, etc, approximately, given the two functions:

01-- The rule for whether or not `year` is a leap year.
02function isleap(year)
03    if year % 400 == 0 then
04        return true;
05    elseif year % 100 == 0 then
06        return false;
07    elseif year % 4 == 0 then
08        return true;
09    end
10    return false;
11end
12 
13function calendar( t )
14    -- t is seconds since beginning of 1970
15    local minutes = t / 60;
16    local hours = minutes / 60;
17    local days = hours / 24;
18    local week = (days % 7 + 4) % 7;
19    -- Weekday; 0 Su, 1 Mo, 6 Sa.
20    -- Year / month is annoying because of leapyears:
21    local dayStart = 0;
22    local year;
23    for y = 1970, 2100 do
24        local len = 365;
25        if isleap(y) then
26            len = 366;
27        end
28        dayStart = dayStart + len;
29        if dayStart > days then
30            year = y;
31            dayStart = dayStart - len;
32            break;
33        end
34    end
35    hours = hours % 24;
36    minutes = minutes % 60;
37    days = days - dayStart;
38    local mos = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
39    local monthStart = 0;
40    if isleap(year) then
41        mos[2] = 29;
42        days = days % 366;
43    else
44        days = days % 365;
45    end
46    for mo = 1, 12 do
47        monthStart = monthStart + mos[mo];
48        if monthStart > days then
49            monthStart = monthStart - mos[mo];
50            month = mo;
51            break;
52        end
53    end
54    local date = days - monthStart;
55    print("Today's Date:");
56    local monthName = {
57        "January", "February", "March", "April", "May", "June", "July",
58        "August", "September", "October", "November", "December"
59    };
60    print(math.floor(date) + 1, monthName[month], year);
61    print(math.floor(hours), ":", math.floor(minutes));
62end
63 
64calendar( os.time() );
65--Today's Date:
66-- 12 December 2014
67-- 19 : 40
68 
69calendar( tick() );
70-- Today's Date:
71-- 12 December 2014
72-- 14 : 40
 View all 72 lines...

os.time() results in the time in GMT (English time).

tick() results in my local time (EST)

Time is bad. If you can, avoid dealing with it. Using HTTP service to query some web service about time information might almost be a better option.

Or, finding a library that deals with it properly (I'm sure there is one, somewhere)

Answered by

adark 5487

10 years ago

os.time is actually a function, not a readable value.

`1`	`print(os.time())` `--seconds passed since the UNIX epoch, UTC`

If you call it with a table of parameters, you can get the seconds since the passed-in time.

Could you give me an example? lightpower26 399 — 10y

It's a *function*. *Methods* are on *instances* of objects and thus usually use the :method() notation. BlueTaslem 18071 — 10y

os.time - How to use it?

2 answers

Answer this question