# C++ output date in number of days per year

for example
user puts in the month as 1-12
the day for whatever month
and the year, which has to include leapyears
leap years only occur when the year is divisible by 4 y%4==o or if divisible by 400 but NOT 100, so 2000 is a leap year but not 2100

so if somebody puts in month 1 day 1 and 2005 it will say 1
if it's month 12 and day is 31 and year is 2005 it will say 365
but if it was 2004 it would say 366
month 3, day 10, year 2005, would be 31+28+10 or just 69

can anybody do this
i guess it needs boolean stuff for the leap year
and a few switch and case things for the rest

i'm not feeling it though

i got one to kind of work with if statements but not correctly, and it was long

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Code:
if (year%4 == 0 && !(year%100 == 0 && year%400 != 0))
DaysPerMonth = {31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
else
DaysPerMonth = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

for (i=0; i<month - 1; i++)
ndays += DaysPerMonth[i];

ndays += day;

Well put :)

and what am i putting in right above that (i've been doing this for less than a month so bear with me)
i dont know what exactly i am calling what
int days, etc.???

pick up a standard C OR C++ book. there are tutorials about such things usual in those or there are websites...sorry I cant remember C++tutorials.com cplus.com i think are some...

yeah i got a book, but it's no good, or it's too good

neurocomp2003 said:
pick up a standard C OR C++ book. there are tutorials about such things usual in those or there are websites...sorry I cant remember C++tutorials.com cplus.com i think are some...
heh I dont think thats a legal URL format, only alpha-numerical chars are allowed :tongue2: