问题描述
大家好，
我需要以秒为单位计算当前时间值：当前时间
是12/16/2003，它'自1970年1月1日以来，整数的时间值应该在1071560616左右.b $ b秒。在Tcl中很容易得到这样的值，但是我注意到它在C中很难做。好吧，ANSI提供了诸如
time（），asctime等功能。 time.h文件中的（），localtime（），gmtime（），ctime（）。但是，
我无法在几秒钟内将回报转换为整数。
我确实想到了计算方式：
年* 365 +月* 12 +天* 24 + .....
来自struct tm成员的值。但我不能处理好闰年和闰月。有人可以帮忙吗？谢谢！
Huey
Hi All,
I need to compute the current time value in seconds: the current time
is 12/16/2003, and it''s time value in integer should around 1071560616
seconds since 1/1/1970. It is easy to get such a value in Tcl, but i
noticed it is uneasy to do in C. Well, ANSI provides functions such as
time(), asctime(), localtime(), gmtime(), ctime() in time.h file. But,
I can''t convert the returns into integer in seconds.
I did think of the way as computing by:
year*365 + month * 12 + day *24 + .....
with the values derived from struct tm members. but I can''t handle
the leap year and leap month well. Anybody can help? Thanks!
Huey
推荐答案
假设您输入的格式为mm / dd / yyyy：
支票，必须/必须扩展
最终处理小时，分钟和秒以及夏令时。
我建议你阅读描述一个结构小心:)：/ b
time_t StringToTimestamp（const char * time_string）
{
struct tm time_in = {0} ;
char * tmp = NULL;
time_in.tm_year = strtoul（time_string + 6，& tmp，10） - 1900;
time_in.tm_mon = strtoul（time_string，& tmp，10） - 1;
time_in.tm_mday = strtoul（time_string + 3，& tmp，10）;
/ *以秒为单位转换日期（自1970年1月1日起）* /
返回mktime（& time_in）;
} / * StringToTimest amp * /
HTH
Robert
Assuming your input is in the format mm/dd/yyyy:
Checks, necessary headers and calling function omitted, may/must be expanded
eventually to handle hours, minutes and seconds as well as daylight saving.
I suggest you read the description of a struct tm carefully :)
time_t StringToTimestamp(const char *time_string)
{
struct tm time_in = {0};
char *tmp = NULL;
time_in.tm_year = strtoul(time_string + 6, &tmp, 10) - 1900;
time_in.tm_mon = strtoul(time_string, &tmp, 10) - 1;
time_in.tm_mday = strtoul(time_string + 3, &tmp, 10);
/* Convert the date in seconds (since 1/1/1970) */
return mktime(&time_in);
} /* StringToTimestamp */
HTH
Robert
这里有一种方式：
#include< stdio.h>
#include< time.h>
#include< stdlib.h>
int main （无效）{
struct tm epoch_strt;
time_t basetime;
time_t curtime;
double nsecs;
epoch_strt.tm_sec = 0;
epoch_strt.tm_min = 0;
epoch_strt.tm_hour = 0;
epoch_strt.tm_mday = 1;
epoch_strt.tm_mon = 1;
epoch_strt.tm_year = 70;
epoch_strt.tm_isdst = - 1;
basetime = mktime（& epoch_strt）;
if（basetime ==（time_t）-1）{
fprintf（stderr，无法转换时间\ n）;
退出（EXIT_FAILURE）;
}
curtime = time（NULL）;
if（curtime ==（time_t）-1）{
fprintf（stderr，无法获取当前时间\ n） ;
退出（EXIT_FAI （LURE）;
}
nsecs = difftime（curtime，basetime）;
printf（" Seconds从纪元开始：％。0f \ n"，nsecs）;
返回0;
}
-nrk。
Here''s one way:
#include <stdio.h>
#include <time.h>
#include <stdlib.h>
int main(void) {
struct tm epoch_strt;
time_t basetime;
time_t curtime;
double nsecs;
epoch_strt.tm_sec = 0;
epoch_strt.tm_min = 0;
epoch_strt.tm_hour = 0;
epoch_strt.tm_mday = 1;
epoch_strt.tm_mon = 1;
epoch_strt.tm_year = 70;
epoch_strt.tm_isdst = -1;
basetime = mktime(&epoch_strt);
if ( basetime == (time_t)-1 ) {
fprintf(stderr, "Cannot convert time\n");
exit(EXIT_FAILURE);
}
curtime = time(NULL);
if ( curtime == (time_t)-1 ) {
fprintf(stderr, "Cannot get current time\n");
exit(EXIT_FAILURE);
}
nsecs = difftime(curtime, basetime);
printf("Seconds since start of epoch: %.0f\n", nsecs);
return 0;
}
-nrk.
假设您输入的格式为mm / dd / yyyy：
检查，必要的标题和调用函数省略，可能/必须最终扩展以处理小时，分钟和秒以及夏令时。我建议你仔细阅读一个struct tm的描述。
time_t StringToTimestamp（const char * time_string）
{struct / time tm time_in = {0};
char * tmp = NULL;
time_in.tm_year = strtoul（time_string + 6，& tmp，10） - 1900;
time_in.tm_mon = strtoul（time_string，& ; tmp，10） - 1;
time_in.tm_mday = strtoul（time_string + 3，& tmp，10）;
/ *以秒为单位转换日期（自1970年1月1日起）* /
返回mktime（& time_in）;
} / * StringToTimestamp * /
HTH
Robert
Assuming your input is in the format mm/dd/yyyy:
Checks, necessary headers and calling function omitted, may/must be
expanded eventually to handle hours, minutes and seconds as well as
daylight saving. I suggest you read the description of a struct tm
carefully :)
time_t StringToTimestamp(const char *time_string)
{
struct tm time_in = {0};
char *tmp = NULL;
time_in.tm_year = strtoul(time_string + 6, &tmp, 10) - 1900;
time_in.tm_mon = strtoul(time_string, &tmp, 10) - 1;
time_in.tm_mday = strtoul(time_string + 3, &tmp, 10);
/* Convert the date in seconds (since 1/1/1970) */
return mktime(&time_in);
} /* StringToTimestamp */
HTH
Robert
mktime的回报是否保证是自纪元以来的秒数？
-nrk。
Is the return of mktime guaranteed to be seconds since epoch?
-nrk.
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