MATLAB: Please Help Basics Made Hard

basics made hard

Hey guys, i am trying to solve this assignemnt. :
Write a function called day_diff that takes four scalar positive integer inputs, month1, day1, month2, day2. These represents the birthdays of two children who were born in 2015. The function returns a positive integer scalar that is equal to the difference between the ages of the two children in days. Make sure to check that the input values are of the correct types and they represent valid dates. If they are erroneous, return -1. An example call to the function would be >> dd = day_diff(1,30,2,1); which would make dd equal 2. You are not allowed to use the built-in function datenum or datetime. Hint: store the number of days in the months of 2015 in a 12-element vector (e.g., 31, 28, 31, 30 …) and use it in a simple formula.
I have figured the Hard part but i have problems how to make it accpent only whole possitve numbers and no matrixes as an input. Here is my code :
function [d] = day_diff( month1,day1,month2,day2 )
   M=[31,28,31,30,31,30,31,31,30,31,30,31];
      if month1==fix(month1)&&month1==fix(month1) && day1==fix(day1) && day2==fix(day2) && isscalar(month1)&&isscalar(day1)&&isscalar(month2)&&isscalar(day2) && day1>0 && day2 > 0 && month1 <=12 && month1 >= 1 && month2 <=12 && month2 >= 1 && M(month1) >= day1 && M(month2) >= day2
         d1 = day1 + sum(M(1:1:month1-1));
         d2 = day2 + sum(M(1:1:month2-1));
         dd = d2-d1;
           if dd <0
               d = dd*(-1);
           else 
               d = dd;
           end
      else d = -1;
      end            
end

Best Answer

if month1 == fix(month1) && month1==fix(month1) ...
This fails, if month1 is not scalar, because than the == comparison replies a logical array and this collides with using the && operator. The solution is easy: Check at first if teh inputs are scalars:
% Short-circuting!
if isscalar(m1) && isscalar(d1) && isscalar(m2) && isscalar(m1) && ...
   m1 == fix(m1) && m1==fix(m1) && d1==fix(d1) && d2==fix(d2) && ...
   d1 > 0 && d2 > 0 && m1 <= 12 && m1 >= 1 && m2 <=12 && m2 >= 1 && ...
   M(m1) >= d1 && M(m2) >= d2
The && operator is "lazy": It does not evaluate the 2nd argument, when the first is false already. This is called "short-circuting".
In this order the terms "M(m1)" and "M(m2)" are secure also: m1 and m2 has been tested before to be legal indices: integer values and in the allowed range.
I've used "m1" instead of "month1" here, because this is easier to read an matches into the interface of the forum. It is questionable if "month1" or "m1" is better.
Hint: With dd = abs(d2 - d1); you do not have to care about the sign.
Note: You've shown your code and your own effort and asked almost a specific question. Therefore I post an answer although I assume this is a homework question. By the way: This problem appeared already in the forum.
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