how to identify leap years
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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
0 Kommentare
Antworten (2)
Siva Charan
am 30 Sep. 2023
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
0 Kommentare
Stijn Haenen
am 18 Mai 2020
Bearbeitet: Stijn Haenen
am 18 Mai 2020
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 Kommentare
Stephen23
am 18 Mai 2020
These are all divisible by four: 1500, 1700, 1800, 1900, 2100, 2200, 2300, 2500, but none of them are leap years.
Steven Lord
am 18 Mai 2020
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