MATLAB：为什么函数无法在给定的日期输入中处理错误？

Question

I'm studying MATLAB on coursera and stuck with this question:

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 (eg, 31, 28, 31, 30 …) and use it in a simple formula.

Multiple cases are tested using a pre-made evaluation code. My code so far:

function answer = day_diff(month1, day1, month2,day2)
    answer = -1;
    days_in_months = [31,28,31,30,31,30,31,31,30,31,30,31]; %days in every month array
    flag1 = days_in_months(month1); %to check if day1 is valid
    flag2 = days_in_months(month2); %to check if day2 is valid
    %Non valid values handing
    if nargin < 4
        error('Must have four arguments');
    end
    if ~isscalar(month1) || month1 < 1 || month1 ~= fix(month1) || month1 > 12
        error('month1 needs to be a positive integer and not greater than 12.');
    end
    if ~isscalar(month2) || month2 < 1 || month2 ~= fix(month2) || month2 > 12        
        error('month2 needs to be a positive integer and not greater than 12.');
    end
    if ~isscalar(day1) || day1 < 1 || day1 ~= fix(day1)|| day1 > flag1        
        error('day1 needs to be a positive integer and a valid date.');
    end
    if ~isscalar(day2) || day2 < 1 || day2 ~= fix(day2)|| day2 > flag2        
        error('day1 needs to be a positive integer and a valid date.');
    end
    %end of error handler
    %to find the age in days
    if (month1 == month2)
        inbetween_days = 0;
        if day1 == day2
            first_last_days = 0;
        elseif day1 < day2
            first_last_days = day2 - day1;
        else
            first_last_days = day1 - day2;
        end
    elseif month1 < month2
            inbetween_days = sum(days_in_months(month1+1:month2-1));
            first_last_days = (days_in_months(month1)-day1) + day2;
        else
            inbetween_days = sum(days_in_months(month2+1:month1-1));
            first_last_days = day1 + (days_in_months(month2)-day2);
    end
     answer = first_last_days +  inbetween_days;
end

so far so good but when it's day_diff(2, 29, 1, 22), the evaluation file gives an error instead of handling the wrong pre-defined input (day1 is 29 and maximum is 28), is the problem in how to return the -1 ?

Problem 4 (day_diff):
        Feedback: Your function performed correctly for argument(s) 1, 30, 2, 1
        Feedback: Your function performed correctly for argument(s) 1, 1, 1, 1
        Feedback: Your function performed correctly for argument(s) 1, 1, 1, 2
        Feedback: Your function performed correctly for argument(s) 1, 2, 1, 1
        Feedback: Your function performed correctly for argument(s) 1, 1, 2, 1
        Feedback: Your function performed correctly for argument(s) 2, 1, 1, 1
        Feedback: Your function performed correctly for argument(s) 1, 31, 2, 1
        Feedback: Your function performed correctly for argument(s) 2, 1, 1, 31
        Feedback: Your function performed correctly for argument(s) 1, 1, 12, 31
        Feedback: Your function performed correctly for argument(s) 2, 1, 3, 1
        Feedback: Your function performed correctly for argument(s) 7, 1, 9, 30
        Feedback: Your program made an error for argument(s) 2, 29, 1, 22

    Your solution is _not_ correct.

Answer 1

Yes you never return -1 which they asked for. Remove the error messages and in those cases simply dont do anything and let your answer stay -1. The following is more lines of code then you had and you can improve upon it but this should pass the test you are failing

    function answer = day_diff(month1, day1, month2,day2)
    answer = -1;
    days_in_months = [31,28,31,30,31,30,31,31,30,31,30,31]; %days in every month array
    flag1 = days_in_months(month1); %to check if day1 is valid
    flag2 = days_in_months(month2); %to check if day2 is valid
    %Non valid values handing
    if nargin < 4
        answer=-1;

    elseif ~isscalar(month1) || month1 < 1 || month1 ~= fix(month1) || month1 > 12
        answer=-1;
    elseif ~isscalar(month2) || month2 < 1 || month2 ~= fix(month2) || month2 > 12        
        answer=-1;
    elseif ~isscalar(day1) || day1 < 1 || day1 ~= fix(day1)|| day1 > flag1        
        answer=-1;
    elseif ~isscalar(day2) || day2 < 1 || day2 ~= fix(day2)|| day2 > flag2        
        answer=-1;
    %end of error handler
    %to find the age in days
    elseif (month1 == month2)
        inbetween_days = 0;
        if day1 == day2
            first_last_days = 0;
        elseif day1 < day2
            first_last_days = day2 - day1;
        else
            first_last_days = day1 - day2;
        end
        answer = first_last_days +  inbetween_days;
    elseif month1 < month2
            inbetween_days = sum(days_in_months(month1+1:month2-1));
            first_last_days = (days_in_months(month1)-day1) + day2;
            answer = first_last_days +  inbetween_days;
    else
            inbetween_days = sum(days_in_months(month2+1:month1-1));
            first_last_days = day1 + (days_in_months(month2)-day2);
            answer = first_last_days +  inbetween_days;
    end

end

Answer 2

I would remove the answer = -1 in the first line and handle all the errors in one if block, to avoid errors in cases that may fall between the elseif blocks:

% Non valid values handing:
if (nargin ~= 4) ||...
        ~all([isscalar(month1) isscalar(month2) isscalar(day1) isscalar(day2)]) ||... % only scalars
        any([month1,month2,day1,day2]<1) ||... % all input greater then zero
        any(fix([month1,month2,day1,day2]) ~= [month1,month2,day1,day2]) ||... % all are integers
        any([month1,month2,day1,day2] > [12 12 flag1 flag2]) % all are valid dates
    answer = -1;
    return;
end
% end of error handler

if the condition is true, the function assigns -1 to answer and goes out (by return ).

In my opinion, this way of error handling is simpler to follow and less error-prone, but I'm not sure that it will solve the original problem.

You could also remove the return statement, and put all the rest of the function in an else block, ie:

if (nargin ~= 4) ||... % all the condition...           
    answer = -1;
else
    % all the calculation...
    answer = first_last_days +  inbetween_days;
end