How can I compute the leap years?

function leapyr = leapyears(startYear, endYear)
%UNTITLED8 Summary of this function goes here
%   Detailed explanation goes here
leapyr = 0;
i = 0;
for i = startYear:4:endYear

        if mod(startYear, 4) == 0 && mod(endYear, 4) == 0 
            fprintf('%i \n', startYear)

        elseif mod(startYear, 400) == 0 && mod(endYear, 400) == 0 
            fprintf('%i \n', startYear)
        else mod(startYear, 100) == 0 && mod(endYear, 100) == 0 
            fprintf('%i \n', startYear)
        end
  end

end

Hey all, I'm struggling to figure out how I'm supposed to supposed to print a list of leap years given two inputs.

The output is supposed to look like this:

a. Example 1.
>> leapyears(2004,2016)
2004
2008
2012
2016
b. Example 2.
>> leapyears(1881,1907)
1884
1888
1892
1896
1904

Please help and thank you.

[KSSV]

There is inbuilt command/ function leapyear.

Answer 1

[Andrei Bobrov]

Please write script - file now1.m:

out = leapyears(2000,2030)
function out = leapyears(startYear,endYear)
    y = (startYear:endYear)';
    x = rem(y,[4,100,400]);
    out = y(x(:,1)==0 & x(:,2) | x(:,3) == 0);
end

use

>> now1
out =
        2000
        2004
        2008
        2012
        2016
        2020
        2024
        2028
>>

Answer 2

[ES]

 function   [leapyears] = leapyears(startyear,endyear)
  leapyears = [];
  for iYr=startyear:endyear
      if mod(iYr,4)==0 && (mod(iYr,100)~=0 || mod(iYr,400)==0)
          leapyears = [leapyears;iYr];
      end
  end
  end

Answer 3

[Mischa Kim]

Aaron, you could do something like

function leapyr = leapyears(startYear, endYear)
%UNTITLED8 Summary of this function goes here % Detailed explanation goes here
leapyr = [];
for year = startYear:endYear
    if mod(year, 4) == 0            % divisible by 4?
        if ~(mod(year, 100) == 0 && mod(year, 400) ~= 0) 
            leapyr = [leapyr,year]; % simply add leap year to vector of leap years
        end
    end
end
end