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 dif

Question

ledinh lam 2016-11-26

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https://ww2.mathworks.cn/matlabcentral/answers/314072-write-a-function-called-day_diff-that-takes-four-scalar-positive-integer-inputs-month1-day1-month

回答： RAMAKANT SHAKYA 2019-2-8

I can't deal it anymore. I have trouble to check an integer number or an array number in order to input argument in this problem .

function dd = day_diff(month1, day1, month2, day2)
if (month1 && day1 >0)&&(month1
    if true
      % code
    end <= 12)&&(month2 && day2 >0) && month2 <= 12
      if isinteger(month1 &&
      if (month1 == 1 && month2 == 1)||(month1 == 3 && month2 == 3)||(month1 == 5 && month2 == 5)||(month1 == 7 && month2 == 7)||(month1 == 8 && month2 == 8)||(month1 == 10 && month2 == 10)||(month1 == 12 && month2 == 12) && day1<=31 && day2<=31
          if day1 == day2
              total1 = day1;
              total2 = day2;
          elseif day1 ~= day2
              total1 = max(day1,day2);
              total2 = min(day1,day2);
          end
      elseif (month1 == 4 && month2 == 4) ||(month1 == 6 && month2 == 6)||(month1 == 9 && month2 == 9)||(month1 == 11 && month2 == 11)  % months have 30 days 
           if day1 == day2 && day1<=30 && day2<30
              total1 = day1;
              total2 = day2;
          elseif day1 ~= day2 && day1<=30 && day2<30
              total1 = max(day1,day2);
              total2 = min(day1,day2);
           else
                dd=-1; return 
           end
      elseif month1 == 1 && month2 ==2 
         total1 = day1;
         total2 = day2+31;
      elseif (month1 == 2 && day1<=28) && month2 == 1
          total1 = day1 + 31;
          total2 = day2;
          elseif (month1 == 2 && day1>28) && month2 == 1
              dd=-1;
              return
      elseif month1 == 1 && month2 == 12
          total1 = day1;
          total2 = day2 + 334;
      elseif month1 == 2 && month2 == 3
          total1 = day1 + 31;
          total2 = day2 + 59;
      elseif month1 == 1  && day1<=31&& month2 == 4
          total1 = day1;
          total2 = day2 + 90;
          elseif month1 == 1 && day1>31 && month2 == 4
              dd=-1; return
       elseif month1 == 11 && day1<=30 && month2 == 12
          total1 = day1 ;
          total2 = day2 + 30;
         elseif month1 == 11 && day1>30 && month2 == 12
             dd=-1;
             return
      elseif month1 == 7 && month2 == 9
          total1 = day1 + 181;
          total2 = day2 + 243;
          elseif month1 == 2 && day1<=28 && month2 == 6 && day2>30 
              dd=-1; return
               elseif month1 == 2 && day1<=28 && month2 == 6 && day2<=30
                 total1 = day1;
                 total2 = day2 + 92;  
      end
  dd = (max(total1,total2)) - (min(total1,total2));
else
dd = -1;
return
end
end

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Answer 1

Image Analyst 2016-11-26

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https://ww2.mathworks.cn/matlabcentral/answers/314072-write-a-function-called-day_diff-that-takes-four-scalar-positive-integer-inputs-month1-day1-month#answer_244772

Why not make your function just be a wrapper for etime()? Did they specifically prohibit using that function?

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Image Analyst 2017-8-26

Floor rounds down to minus infinity. Fix rounds towards zero. They give different results when operating on negative numbers.

champions2015 2017-8-26

I see, thanks very much!

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Answer 2

Ugur Ulas 2017-12-29

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编辑：Ugur Ulas 2017-12-29

在 MATLAB Online 中打开

Here there is more basic solution:

function dd = day_diff(m1, d1, m2, d2)
  A = [31 28 31 30 31 30 31 31 30 31 30 31]';
  day1 = d1 + sum(A(1:(m1-1)));
  day2 = d2 + sum(A(1:(m2-1)));
  if prod(size(m1)) ~= 1 || prod(size(m2)) ~= 1 || prod(size(d1)) ~= 1 || prod(size(d2)) ~= 1
      dd = -1;
  elseif  m1 < 1 || m2 < 1 || d1 < 1 || d2 < 1 || m1 ~= floor(m1) || m2 ~= floor(m2) || d1 ~= floor(d1) || d2 ~= floor(d2)
      dd = -1;
  elseif A(m1) < d1 || A(m2) < d2
      dd = -1;
  else
  dd = abs(day2-day1);
  end
end

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Raunil Raj 2018-3-7

在 MATLAB Online 中打开

Thank you for you response but I actually wanted to know the logic behind this:

elseif A(m1) < d1 || A(m2) < d2
          dd = -1;

Image Analyst 2018-3-8

It says to set dd equal to minus one if A is less than d1 at the m1 index, or if A is less than d2 at the m2 index.

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Answer 3

Vignesh M 2018-5-4

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if true
  function [ dd ] = day_diff( m1,d1,m2,d2 )

% Short circuiting!

N = [31,28,31,30,31,30,31,31,30,31,30,31];

if isscalar(m1) && isscalar(d1) && isscalar(m2) && isscalar(d2) && ... m1 == fix(m1) && d1 == fix(d1) && m2 == fix(m2) && d2 == fix(d2) &&... m1 > 0 && m1 <= 12 && m2 > 0 && m2 <= 12 && ... d1 > 0 && d1 <= N(m1) && d2 > 0 && d2 <= N(m2)

      day1 = d1 + sum(N(1:1:m1-1));
      day2 = d2 + sum(N(1:1:m2-1));
      dd = abs(day1-day2)
else
    dd = -1
end
end

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Answer 4

RAMAKANT SHAKYA 2019-2-8

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https://ww2.mathworks.cn/matlabcentral/answers/314072-write-a-function-called-day_diff-that-takes-four-scalar-positive-integer-inputs-month1-day1-month#answer_360109

在 MATLAB Online 中打开

function dd=day_diff(m1,d1,m2,d2)
      num_of_days=[31,28,31,30,31,30,31,31,30,31,30,31];
      dd=-1;
      mmm1=round(m1);mmm2=round(m2);ddd1=round(d1);ddd2=round(d2);
      if(isscalar(d1)&&isscalar(d2)&&isscalar(m1)&&isscalar(m2)) % checking for valid input
       if(d1-ddd1 || d2-ddd2 || m1-mmm1 || m2-mmm2)  % checking for valid input
             dd=-1;
       else
         if (m1<1||m1>12||m2<1||m2>12) %checking months
                  dd=-1;
              else
                  if(d1>num_of_days(m1) || d2>num_of_days(m2) || d1<1 || d2<1) %validating days of respective month
                      dd=-1;
                  else
                      dd1=sum(num_of_days(1:m1));
                      dd1=dd1+d1-num_of_days(m1);
                      dd2=sum(num_of_days(1:m2));
                      dd2=dd2+d2-num_of_days(m2);
                      if(dd1>=dd2)
                          dd=dd1-dd2;
                      else
                          dd=dd2-dd1;
                      end
                  end
         end
       end        
    end