how to identify leap years

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Anisia a
Anisia a 2020 年 5 月 18 日
回答済み: Siva Charan 2023 年 9 月 30 日
Hi,
I am trying to create a function that takes 3 +ve integer scalar inputs year, month and date. If these 3 represent a valid date return true otherwise return false. My code is running for most of the input. But I am having problem with leap years. The code I wrote is given below. Can anyone please point out my mistake.
function valid= valid_date(year, month, date)
v=1;
a= mode(year,4);
b= mode(year,100);
c= mode(year,400);
if ~isscalar(year) || year<1 || year~=fix(year) || ~isscalar(month) || month<1 || month~=fix(month) ||~isscalar(date)|| date<1 || date~=fix(date)
v= 0;
end
if v==0
valid=false;
return
end
if 0>month || month>12
v= 0;
elseif (month==1 ||month==3|| month==5 || month==7 || month==8|| month==10|| month==12)
if 0<date&& date<=31
v=1;
else
v=0;
end
elseif (month== 4 ||month==6|| month==9 || month==11)
if 0<date&& date<31
v=1;
else
v=0;
end
elseif month==2
if date==29
if (a==0 && b~=0) || c==0
v=1;
else
v=0;
end
elseif 0<date && date<29
v=1;
else
v=0;
end
end
if v==0
valid=false;
else
valid=true;
end

回答 (2 件)

Siva Charan
Siva Charan 2023 年 9 月 30 日
Use this subfunction and call from the main function.
if the year is divisible by 100 and not divisible by 400, it is not a leap year. millennium years(1600, 1700, 1800...) should be divisible by 400 to be leap years, for others, any year that is divisible by 4 can be a leap year.
function leapyear = checkleap(year)
if mod(year,100)==0
if mod(year,400)==0
leapyear = true;
else
leapyear = false;
end
elseif mod(year,4) == 0
leapyear = true;
else
leapyear = false;
end
end

Stijn Haenen
Stijn Haenen 2020 年 5 月 18 日
編集済み: Stijn Haenen 2020 年 5 月 18 日
There is a leap year every four years, so you can use this:
if mod(year,4)==0
'leap year'
else
'not a leap year'
end
  4 件のコメント
Stephen23
Stephen23 2020 年 5 月 18 日
These are all divisible by four: 1500, 1700, 1800, 1900, 2100, 2200, 2300, 2500, but none of them are leap years.
Steven Lord
Steven Lord 2020 年 5 月 18 日
Yes. The Wikipedia page has the algorithm for how to determine whether a year is a leap year.

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