Ohm v2, 5 bytes

υ!+υy 

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Explanation:

υ!+υy υ! get the current timestamp in seconds + sum with the implicit input υy get the year from the resulting timestamp 

Bash, 16 bytes

date -d$1sec +%Y 

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Credits

• Saved 12 bytes thanks to @manatwork

PHP, 27 bytes

<?=date(Y,time()+$argv[1]); 

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05AB1E, 116 bytes

"`т‰0Kθ4ÖUD2Qi\28X+ë<7%É31α}"Vžežfžg)IŽª+·÷ÄF©IdiY.V‹i®¬>0ë®1¾ǝDÅsD12‹i>1ë\1Dǝ¤>2}}ǝëć©ić©i<D_-12š31šë®<šDY.Všë®<š]θ 

Here we go again..

Try it online or try it only with custom specified starting date. (Pretty slow for large inputs, so won't be able to output the larger test cases with the current date.)

Explanation:

Since 05AB1E doesn't have any date builtins (except for the current day/time), I've calculated things manually before. I've used the code of going to the next day from this answer of mine, which in turns also uses the leap year calculation of this answer of mine.
Since this challenge also asks to go back in time, I've modified the program accordingly to support that as well.

Step 1: Create a function to calculate the amount of days in a month for a given year/month:

"`т‰0Kθ4ÖUD2Qi\28X+ë<7%É31α}" # Push this string V # Pop and store it in variable `Y` ` # Pop the list and push all values separated to the stack # (the year will be at the top; month below it) т‰ # Take the divmod-100 of the year 0K # Remove the 0s θ # Pop and push its last item 4Ö # Check if it's divisible by 4 U # Pop and store this in variable `X` # (X=1 for leap years; X=0 for non-leap years) D # Duplicate the current month 2Qi # If it's equal to 2 (thus February): \ # Discard the duplicated month 28X+ # Push 28 + isLeapYear from variable `X` ë # Else: < # Decrease the month by 1 7% # Modulo-7 É # Modulo-2 31α # Absolute difference with 31 } # Close the if-else statement 

Step 2: Now we determine the current date, and loop an amount of times depending on the input:

že # Push the current day žf # Push the current month žg # Push the current year ) # Wrap all three into a list I # Push the input-integer Žª+ # Push compressed integer 42300 · # Double it to 84600 (60*60*24) ÷ # Integer-divide the input by 84600 Ä # Take its absolute value F # And loop that many times: 

Step 3: If the input was positive, calculate the next date:

© # Store the current date in variable `®` (without popping) Idi # If the input is non-negative: Y.V # Execute string `Y` as 05AB1E code to get the amount # of days for the current month/year ‹i # If this is smaller than the current days: ® # Push the current date again ¬ # Get its first item (the days) > # Increase the day by 1 0 # Push index 0 ë # Else: ® # Push the current date again 1 # Push day=1 ¾ # Push index 0 ǝ # Insert days=1 at index 0 into the date D # Duplicate it Ås # Pop and push its middle item (the month) D # Duplicate that month 12‹i # If it's smaller than 12: > # Increase the month by 1 1 # Push index 1 ë # Else: \ # Discard the duplicated month 1 # Push month=1 D # Push index 1 ǝ # Insert month=1 at index 1 into the date ¤ # Push its last item (the year) > # Increase the year by 1 2 # Push index 2 } # Close the if-else statement } # Close the if-else statement ǝ # Insert the value at the given index into the date 

Step 4: If the input was negative instead, calculate the previous date:

ë # Else (the input is negative): ć # Extract head; push [month,year] and day separated © # Store the day in variable `®` (without popping) i # If day == 1: ć # Extract head; push [year] and month separated © # Store the month in variable `®` (without popping) i # If month == 1: < # Decrease the [year] by 1 D # Duplicate it _ # Check if it's equal to 0 (1 if 0; 0 otherwise) - # Subtract that (so we go from year 1 to -1) 12š31š # Prepend 12 and 31: [31,12,year-1-(year-1==0)] ë # Else: ®<š # Prepend month-1 to the [year] list D # Duplicate it Y.V # Execute string `Y` as 05AB1E code to get the amount # of days for this month/year š # Prepend this to the [month-1,year] list ë # Else: ®<š # Prepend day-1 to the [month,year] list 

Step 5: After the loop, extract the year to output:

] # Close both the if-else and loop θ # Pop the date, and only leave its last item (the year) # (after which it is output implicitly as result) 

See this 05AB1E tips of mine (section How to compress large integers?) to understand why Žª+ is 42300.

Ruby, 21 bytes

->n{[*Time.now+n][5]} 

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Shorter by a byte than the more obvious ->n{(Time.now+n).year}.

Octave / MATLAB, 27 bytes

@(x)datestr(x/86400+now,10) 

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How it works

@(x) % Define anonymous function with input x x/86400 % Divide input by 86400, to convert to days now % Current time in days since "January 0, 0000" + % Add datestr( ,10) % Convert to string with format 10, which is year 

JavaScript (ES6), 42 bytes

n=>new Date(+new Date+n*1e3).getFullYear() 

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jq, 20 15 characters

now+.|gmtime[0] 

Sample run:

bash-5.0$ jq 'now+.|gmtime[0]' <<< 100000000000 5190 

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PowerShell, 32 26 23 bytes

date|% *dds* @args|% y* 

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-3 bytes thanks to mazzy and ZaelinGoodman

Python 3, 74 72 65 bytes

lambda x:(date.today()+timedelta(0,x)).year from datetime import* 

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-7 bytes thanks to @Eric

Japt, ¹² 10 bytes

ÐKj +Ue3)i 

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-2 bytes from AZTECCO.

This transpiles to the following JS:

new Date(K.j() + U * 1e3).i() 

where K.j() is the current time converted to milliseconds, U is the input, and .i() gets the year from the date constructor.

R, 12 bytes

Sys.time()+n 

This returns a full date, the year is at the beginning of the string. This is according to the rules, which didn't say to print just the year; and this is not in the list of "standard loopholes" by the time of this answer.

To print just a year, R would need 25 bytes:

format(Sys.time()+n,"%Y") 

IBM/Lotus Notes Formula Language, 34 32 Bytes

@Year(@Adjust(@Now;0;0;0;0;0;i)) 

Takes input from a field named i. Unfortunately, Formula only supports 32 bit signed integers so it is restricted to a maximum input of +/- 2,147,483,647.

There is no TIO for Formula so here are a couple of screenshots:

MATL, 11 bytes

12L/Z'+10XO 

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Explanation

 % Implicitly grab input as an integer (seconds) 12L % Push the literal 86400 to the stack (seconds in a day) / % Divide the input (in seconds) by the seconds in a day to convert to days Z' % Get the current time in days as a float + % Add the current time and the seconds in the future together 10XQ % Determine the year of this date % Implicitly display the result 

Julia, 38 bytes

using Dates f(x)=Year(now()+Second(x)) 

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Raku, 28 bytes

{DateTime.new($_+time).year} 

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Python 3, 51 bytes

lambda x:gmtime(time()+x).tm_year from time import* 

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AWK, 30

$1=strftime("%Y",$1+systime()) 

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• 5 bytes saved thanks to @manatwork.

GNU coreutils (16 bytes)

date +%Y -d$1sec 

Saved 4 bytes, thanks to @ovs.

MySQL, 35 bytes

select year(now()+interval [input] second)

Java (JDK), 57 bytes

n->java.time.LocalDateTime.now().plusSeconds(n).getYear() 

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Beats the Date API (71 bytes):

n->new java.util.Date(System.currentTimeMillis()+n*1000).getYear()+1900 

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Note that this one returns the year in absolute form (no negative year).

And I don't even speak about the Calendar API.

C# (.NET Core), 98 bytes

public class E{public static void m(double x){Console.WriteLine(DateTime.Now.AddSeconds(x).Year);} 

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Didn't count the using System; and calling function in Main(), if they needed to be added, please inform me!

Visual Basic .NET (VBC), 89 bytes

public class E Shared sub m(x) Console.WriteLine(DateTime.Now.AddSeconds(x).Year) end Sub 

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VBScript, 52 bytes

MsgBox Year(DateAdd("s",Wscript.Arguments(0),Now()) 

C# (Visual C# Interactive Compiler), 34 bytes

x=>DateTime.Now.AddSeconds(x).Year 

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AppleScript function, 46

to y(s) (current date+s/60*minutes)'s year end 

Test driver:

repeat with sec in {100000000000, 1000, 54398643, 54398643234, -100000000000} log y(sec) end repeat 

Save the function and the test driver as text to nseconds.applescript, then run in the terminal:

osascript nseconds.applescript 

Or run in the AppleScript Editor app.

T-SQL, 42 bytes

select year(dateadd(s,a,getdate()))from t; 

Uses the standard input method for T-SQL of taking input from a table. You can create a suitable table like this:

create table t (a int not null); insert into t values (1000000000), (1000), (54398643), (-1000000000); 

Note that T-SQL is limited to 32-bit integers for the number of seconds. To avoid that and allow the resulting year to vary between 0 and 9999 costs 6 bytes:

select year(dateadd(d,a/86400,getdate()))from t; 

Don't forget to declare a as bigint if you do this. (Note that hours is represented by hh so it's no shorter, and minutes won't let you reach 9999.)

APL (Dyalog Unicode), 20 bytes (SBCS)

Full program, prompting for n from stdin.

⊃⊃20 ¯1⎕DT⎕+20⎕DT'J' 

20⎕DT'J' get current local* (Juliett) DateTime as UNIX time (code 20; seconds from beginning of 1970)

⎕+ prompt for n and use that to increment the UNIX time.

20 ¯1 interpret as UNIX times (code 20) and convert to an array of time stamps (code -1)

⊃ extract the first (and only) time stamp

⊃ extract the first element (the year)

Try APL! (first and last case scaled down by a factor of 10 to avoid hitting date-time limits, 1 January 0001 though 28 February 4000; turned into function for ease of use)

* could also have 'Z' (Zulu) to use the current UTC time instead of local time.

Perl 5 -p, 23 bytes

$_=gmtime$^T+$_;s/.* // 

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Mathematica, 43 bytes

IntegerPart@DatePlus[#/86400]["YearExact"]& 

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Thanks to @LegionMammal978 for hinting that DatePlus[#/86400][[1,1]]& fails for negative years