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gregorian calendar, help

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gregorian calendar, help

 
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  #1
Mar 21st, 2008
Hello Geeks,
I am working on a program for date,month calculation.
I found this code used to convert to Gregorian format for calculation.
It would be great full for me if anybody can explain this ... like 0.75 yy/4 .4*mm what are these constants why m-1, yy-1 etc..
C Syntax (Toggle Plain Text) 
  1. long xx;
  2. int yy,dd,m;
  3. xx = 365 + yy +dd+31*(m-1)
  4. if(m<3)
  5. xx= xx+((yy-1)/4)-(0.75*(yy-1)/100 + 1);
  6. else
  7. xx = xx-(0.4*mm+2.3)+(yy/4)-(0.75*(yy/100)+1));
Thankyou
Last edited by modaslam; Mar 21st, 2008 at 7:58 am.
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Re: gregorian calendar, help

 
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Mar 21st, 2008
>what are these constants
Why don't you ask the person who created the formula?

>why m-1, yy-1 etc..
I imagine because m and yy are 1 based, and the formula works better with values that are 0 based.
I'm here to prove you wrong.
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Re: gregorian calendar, help

 
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Mar 21st, 2008
I have no idea why the original coder wrote that stuff, except it appears the s/he did not know about the standard C library date/time functions.

date arithmetic is best done with that library - convert a date/time to a struct tm, convert the struct tm to epoch seconds. Do the same thing with another time or date. Now you can add subtract or whatever - like what date & time was it 48 days ago? You don't have to worry about leap years, for example.

Work backwards to get a new date.

The functions you need are
gmtime or localtime
mktime
strftime

You also have to learn what goes into struct tm, so you can build one if your system does not have strptime.
here is some help:
http://publications.gbdirect.co.uk/c..._and_time.html
Last edited by jim mcnamara; Mar 21st, 2008 at 2:27 pm.
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Re: gregorian calendar, help

 
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Mar 22nd, 2008
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Originally Posted by jim mcnamara View Post
I have no idea why the original coder wrote that stuff, except it appears the s/he did not know about the standard C library date/time functions.

date arithmetic is best done with that library - convert a date/time to a struct tm, convert the struct tm to epoch seconds. Do the same thing with another time or date. Now you can add subtract or whatever - like what date & time was it 48 days ago? You don't have to worry about leap years, for example.

Work backwards to get a new date.

The functions you need are
gmtime or localtime
mktime
strftime

You also have to learn what goes into struct tm, so you can build one if your system does not have strptime.
here is some help:
http://publications.gbdirect.co.uk/c..._and_time.html
Its not like that. we are not supposed to use any library functions.. for finding the date.
Actually the program is to find the days elapsed between to given days.

The code snippet i have given is to convert the normal date to Gregorian type.

what they have explained is first they take two days and convert to Gregorian type.
and subtract second from first. so that they can get the difference in days elapsed

To convert a normal date to Gregorian type they used this function.
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Re: gregorian calendar, help

 
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Mar 23rd, 2008
I am wondering if we can build a calendar in C language...

...as in Gregorian Calendar format and design

Is there any code that could help me to print or to display a Gregorian Calendar???

pls. reply... thanks... God bless...


my yahoo email add is: email snipped
Last edited by WolfPack; Mar 24th, 2008 at 12:16 pm. Reason: Removed email address per forum rules.
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Re: gregorian calendar, help

 
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Mar 24th, 2008
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Originally Posted by jonnelgrande View Post
I am wondering if we can build a calendar in C language...

...as in Gregorian Calendar format and design

Is there any code that could help me to print or to display a Gregorian Calendar???

pls. reply... thanks... God bless...


my yahoo email add is: email snipped
The code i have given in this thread accept a normal date of format dd mm yy format and convert to Gregorian format well basically the this snippet is function used to convert the date.
okay
Last edited by WolfPack; Mar 24th, 2008 at 12:17 pm. Reason: Removed email from quoted message.
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