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1. The problem statement, all variables and given/known data:
This function is given the day, month and year and returns the Julian date. The Julian date is the ordinal day number for that day. For example, 1 Jan is day 1 of any year, 31 Dec is day 365 for any non-leap year and 1 Feb is day 32 for any year. Use your days_in_month() function from the previous problem to calculate the Julian date. Put the code for the function julian_date() in the file julian.c, and the prototype in the file julian.h Write a driver, main(), which asks the user to enter a month, day, year and prints the Julian date. Terminate with EOF. Put the driver in the file driver3.c. You will compile this program with the command:
2. Relevant commands, code, scripts, algorithms:
Other code that relates to it would be the days.h and leap.h
3. The attempts at a solution (include all code and scripts):
for julian.c 4. Complete Name of School (University), City (State), Country, Name of Professor, and Course Number (Link to Course):
University of Hawaii at Manoa, Honolulu (HI), Oahu (Hawaii), Tep Dobry, EE 160
( i can't post any URL for i am new to this forum but type tep dobry in google and the link to the course should be titled with 2011 after the link is connected the information is in its in the homework section)
If you need more info i'll post up more
Thank you for your time
Note: Without school/professor/course information, you will be banned if you post here! You must complete the entire template (not just parts of it).
Last edited by Scott; 03-06-2011 at 06:40 PM..
Reason: Added code tags
First, I must say that year professor has thrown you a curve. As this page says:
Quote:
The use of Julian date to refer to the day-of-year (ordinal date) is usually considered to be incorrect although it is widely used that way in the earth sciences, computer programming, and the food industry.
The description of the algorithm given is clearly day-of-year, not a real Julian Day. This may confuse you as you research the subject.
I won't write this for you, but a few comments on your code... year==leap That can't be right. Even if there was only one leap year, you never set the variable called leap to anything. You need to calculate whether a year is a leap year or not. A good algorithm is on the Wikipedia page for leap year. And remember that C has a % to indicate modulus. So the expression year%400 is the remainder when year is divided by 400. If this is zero then year was an exact multiple of 400.
In January,the day-of-year and the day-of-month are the same. Why do you have day=day+1?
I don't see any driver... What this means is a main program that calls your function and returns the result.
I also don't see your days_in_month() function mentioned in the problem. Ideally I think it should take both a year and and a month because February changes in a leap year. But it may have been defined differently.
I think there are enough clues here for you to take another crack at this routine. Good luck!
Last edited by Perderabo; 03-06-2011 at 07:49 PM..
Reason: Correct grammer
How to get Julian date (Three digit) of a given date (Not current date)? I do not have root privilege - so can not use date -d. Assume that we have three variables year, month and date.
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use Time::Local;
use POSIX qw(strftime);
my $time=timelocal(1,2,3,9,11,2013);
printf strftime "%j", localtime($time);
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Hi all,
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----------------------------------------
gr8 if give very small command/script
and please explain the steps as well(imp)
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im new for UNIX. i have a problem in date function. please help me to find a solution.
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Hi,
Was using date +%Y%j to get current julian date. Can anyone let me know how can I get y'day's julin date. Thx
Did check FAQ but couldn't find anything.
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I have script in unix which creates a julian date like 126 or 127
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srikanth (6 Replies)
Hi,
is there any possibility to find julian date for given corresping date.
I will be gladfull if i get it.
Requirement :
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