MATLAB: Convert to a function

I would like this code to be converted to a function.
% Convert to function
year=input('Enter specified year(yyyy):');
if year>0
  if mod(year,400)==0
      leap_day=1;
  else if mod(year,100)==0
          leap_day=0;
      else if mod(year,4)==0
              leap_day=1;
          else 
              leap_day=0;
          end
      end
  end
end
          month =input('Enter specified month(1-12):');
          if month>=1 && month <=12
              switch (month)
                  case {1,3,5,7,8,10,12}
                      max_day=31;
                  case {4,6,9,11}
                      max_day=30;
                  case {2}
                      max_day=28+leap_day;
              end
              fprintf('Enter specified day (1-%d):',max_day);
              day=input('');
              if day>=1 && day <=max_day
                  day_of_year=day;
                  for ii=1:month-1
                      switch(ii)
                          case {1,3,5,7,8,10,12}
                              day_of_year=day_of_year+31;
                          case {4,6,9,11}
                              day_of_year=day_of_year+30;
                          case {2}
                              day_of_year=day_of_year+28+leap_day;
                      end
                  end

Best Answer

>> calender1 = @(day1,month1,year1)datenum(year1,month1,day1) - datenum(year1 - 1,12,31);
>> calender1(3,1,2017)
ans =
     3
>> calender1(3,3,2020)
ans =
    63
>> calender1(3,3,2017)
ans =
    62
>>
or create m - file: calcDaysOfYear.m
function num_of_day = calcDaysOfYear(day1,month1,year1)
    dm = 30*ones(12,1);
    x = rem(year1,[4,100,400]);
    dm(2) = dm(2) - 1 - (x(:,1) | x(:,2)== 0 & x(:,3));
    dm([1:2:7,8:2:end]) = dm([1:2:7,8:2:end]) + 1;
    num_of_day = day1 + sum(dm(1:month1-1));
end
use
>>calcDaysOfYear(3,1,2017)
  ans =
       3
>>calcDaysOfYear(3,3,2016)
  ans =
       63
>>calcDaysOfYear(3,3,2017)
  ans =
       62
Related Question