local sakamoto = {0, 3, 2, 5, 0, 3, 5, 1, 4, 6, 2, 4};
local function is_leap ( y )
- return (y % 4) == 0 and (y % 100) ~= 0 or (y % 400) == 0
+ if (y % 4) ~= 0 then
+ return false
+ elseif (y % 100) ~= 0 then
+ return true
+ else
+ return (y % 400) == 0
+ end
end
local function year_length ( y )
return ( year + leap_years_since ( year ) + sakamoto[month] + day ) % 7 + 1
end
-local function increment ( tens , units , base )
+local function borrow ( tens , units , base )
+ local frac = tens % 1
+ units = units + frac * base
+ tens = tens - frac
+ return tens , units
+end
+
+local function carry ( tens , units , base )
if units >= base then
tens = tens + idiv ( units , base )
units = units % base
-- Modify parameters so they all fit within the "normal" range
local function normalise ( year , month , day , hour , min , sec )
- min , sec = increment ( min , sec , 60 ) -- TODO: consider leap seconds?
- hour , min = increment ( hour , min , 60 )
- day , hour = increment ( day , hour , 24 )
-
- while day <= 0 do
+ -- `month` and `day` start from 1, need -1 and +1 so it works modulo
+ month , day = month - 1 , day - 1
+
+ -- Convert everything (except seconds) to an integer
+ -- by propagating fractional components down.
+ year , month = borrow ( year , month , 12 )
+ -- Carry from month to year first, so we get month length correct in next line around leap years
+ year , month = carry ( year , month , 12 )
+ month , day = borrow ( month , day , month_length ( floor ( month + 1 ) , year ) )
+ day , hour = borrow ( day , hour , 24 )
+ hour , min = borrow ( hour , min , 60 )
+ min , sec = borrow ( min , sec , 60 )
+
+ -- Propagate out of range values up
+ -- e.g. if `min` is 70, `hour` increments by 1 and `min` becomes 10
+ -- This has to happen for all columns after borrowing, as lower radixes may be pushed out of range
+ min , sec = carry ( min , sec , 60 ) -- TODO: consider leap seconds?
+ hour , min = carry ( hour , min , 60 )
+ day , hour = carry ( day , hour , 24 )
+ -- Ensure `day` is not underflowed
+ -- Add a whole year of days at a time, this is later resolved by adding months
+ -- TODO[OPTIMIZE]: This could be slow if `day` is far out of range
+ while day < 0 do
year = year - 1
day = day + year_length ( year )
end
+ year , month = carry ( year , month , 12 )
- -- Lua months start from 1, need -1 and +1 around this increment
- month = month - 1
- year , month = increment ( year , month , 12 )
- month = month + 1
-
- -- This could potentially be slow if `day` is very large
+ -- TODO[OPTIMIZE]: This could potentially be slow if `day` is very large
while true do
- local i = month_length ( month , year )
- if day <= i then break end
+ local i = month_length ( month + 1 , year )
+ if day < i then break end
day = day - i
month = month + 1
- if month > 12 then
- month = 1
+ if month >= 12 then
+ month = 0
year = year + 1
end
end
+ -- Now we can place `day` and `month` back in their normal ranges
+ -- e.g. month as 1-12 instead of 0-11
+ month , day = month + 1 , day + 1
+
return year , month , day , hour , min , sec
end
end
local function new_from_timestamp ( ts )
+ if type ( ts ) ~= "number" then
+ error ( "bad argument #1 to 'new_from_timestamp' (number expected, got " .. type ( ts ) .. ")" , 2 )
+ end
return new_timetable ( 1970 , 1 , 1 , 0 , 0 , ts )
end
return {
+ is_leap = is_leap ;
day_of_year = day_of_year ;
day_of_week = day_of_week ;
normalise = normalise ;