日期算术 [英] Date Arithmetic
问题描述
是否有任何本地化的设施用于执行日期算术
是否正确考虑了闰年，夏令时，以及其他什么
格里高利农历的其他特性？ />
具体来说，考虑到以下程序并且它的输出低于
有人可以为time_subtract函数提供合适的主体吗？
Mike
- 8< - program
#include< stdlib.h>
#include< stdio.h>
#include< string.h>
#include< time.h>
time_t *
time_subtract（time_t * timep，struct tm * adj）
{
/ * ???? * /
}
int
main（int argc，char * argv []）
{
struct tm adj;
time_t now;
time_t * time3hoursago;
memset （& adj，0，sizeof adj）;
adj.tm_hour = 3; / * 3小时，而不是第3小时* /
now = time（NULL）;
time3hoursago = time_subtract（& now，& adj）;
puts（ctime（time3hoursago））;
返回EXIT_SUCCESS;
}
- 8< - 输出
$ date
Sat Nov 1 02:36:09 EST 2003
$ ./time_sub
Fri Oct 31 23:36:09 2003
Are there any localized facilites for performing date arithmetic that
properly considers leap years, daylight savings time, and whatever
other peculiarities of the gregorian lunar calendar?
Specifically, considering the following program and it''s output below
can someone provide a suitable body for the time_subtract function?
Mike
--8<-- program
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <time.h>
time_t *
time_subtract(time_t *timep, struct tm *adj)
{
/* ???? */
}
int
main(int argc, char *argv[])
{
struct tm adj;
time_t now;
time_t *time3hoursago;
memset(&adj, 0, sizeof adj);
adj.tm_hour = 3; /* 3 hours, not 3rd hour */
now = time(NULL);
time3hoursago = time_subtract(&now, &adj);
puts(ctime(time3hoursago));
return EXIT_SUCCESS;
}
--8<-- output
$ date
Sat Nov 1 02:36:09 EST 2003
$ ./time_sub
Fri Oct 31 23:36:09 2003
推荐答案
date
11月1日星期六02 ：2003年东部时间36:09
date
Sat Nov 1 02:36:09 EST 2003
./ time_sub
Fri Oct 31 23:36:09 2003
./time_sub
Fri Oct 31 23:36:09 2003
" Michael B Allen" < MB ***** @ ioplex.com>在消息中写道
news：pa ********************************** @ ioplex .c om ...
"Michael B Allen" <mb*****@ioplex.com> wrote in message
news:pa**********************************@ioplex.c om...
是否有任何本地化设施用于执行日期算术，以确定正确考虑闰年，夏令时以及格里高利农历的其他特性？
具体来说，考虑到下面的程序，它的输出可以在下面输出
有人为time_subtract函数提供合适的主体吗？
迈克
- 8< - program
#include< stdlib.h>
#include< stdio.h>
#include< string.h> ;
#include< time.h>
time_t *
time_subtract（time_t * timep，struct tm * adj）
{
/ *？ ??? * /
time_t t = mktime（adj）;
返回difftime（* timep，t）}
int
main（int argc，char * argv []）
{
struct tm adj;
time_t now;
time_t * time3hoursago;
memset（& adj，0，sizeof adj。;
adj.tm_hour = 3; / * 3小时，而不是第3小时* /
现在=时间（NULL）;
time3hoursago = time_subtract（& now，& adj）;
puts（ctime（time3hoursago））;
返回EXIT_SUCCESS;
}
- 8< - 输出
Are there any localized facilites for performing date arithmetic that
properly considers leap years, daylight savings time, and whatever
other peculiarities of the gregorian lunar calendar?
Specifically, considering the following program and it''s output below
can someone provide a suitable body for the time_subtract function?
Mike
--8<-- program
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <time.h>
time_t *
time_subtract(time_t *timep, struct tm *adj)
{
/* ???? */ time_t t = mktime(adj);
return difftime(*timep, t) }
int
main(int argc, char *argv[])
{
struct tm adj;
time_t now;
time_t *time3hoursago;
memset(&adj, 0, sizeof adj);
adj.tm_hour = 3; /* 3 hours, not 3rd hour */
now = time(NULL);
time3hoursago = time_subtract(&now, &adj);
puts(ctime(time3hoursago));
return EXIT_SUCCESS;
}
--8<-- output
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