在没有夏令时转换的情况下将纪元时间转换为人类日期时间字符串

Question

我是 C 的新手，我目前正在从事一个项目，我需要将纪元时间转换为人工日期和时间字符串。

我有 1411210794 的 UTC 时间戳，用于 20th September 2014 10:59:54。

然后我在 UTC 时间戳上运行以下函数以转换为人类可读的时间戳：

void UTCoTimeDate(char** date, char** time, long UTC, BOOL includeSeconds)
{
     time_t epch = UTC;
     struct tm *timeptr =  gmtime(&epch);
     asprintf(date, "%d-%.2d-%.2d", 1900 + timeptr->tm_year, timeptr->tm_mon+1, timeptr->tm_mday);
     if (includeSeconds)
     {
        asprintf(time, "%.2d:%.2d:%.2d", timeptr->tm_hour, timeptr->tm_min, timeptr->tm_sec);
     }
     else
     {
        asprintf(time, "%.2d:%.2d", timeptr->tm_hour, timeptr->tm_min); 
     }
}

当我通过 gdb 运行它并查看 tm *timeptr 内容时，小时已变为 11 而不是停留在 10。以下是 tm 结构的完整内容

{tm_sec = 54, tm_min = 59, tm_hour = 11, tm_mday = 20, tm_mon = 8, tm_year = 114, tm_wday = 6, tm_yday = 262, tm_isdst = 1, tm_gmtoff = 3600,
  tm_zone = 0x806c488 "BST"}

我期待人类约会回归 10:59:54 而不是 11:59:54。

Answer 1

这是一个演示程序(ttt.c)：

#include <stdio.h>
#include <time.h>

static void dump_tm(const char *tag, const struct tm *tm)
{
    printf("%s: {tm_sec = %.2d, tm_min = %.2d, tm_hour = %.2d, tm_mday = %.2d,"
           " tm_mon = %.2d, tm_year = %.3d, tm_wday = %d, tm_yday = %.3d,"
           " tm_isdst = %d, tm_gmtoff = %.5ld, tm_zone = \"%s\"}\n",
           tag, tm->tm_sec, tm->tm_min, tm->tm_hour, tm->tm_mday, tm->tm_mon,
           tm->tm_year, tm->tm_wday, tm->tm_yday, tm->tm_isdst, tm->tm_gmtoff,
           tm->tm_zone);
}

int main(void)
{
    time_t  t0 = 1411210794;
    struct tm *ut = gmtime(&t0);
    struct tm *lt = localtime(&t0);

    dump_tm("UTC", ut);
    dump_tm("Local", lt);
    return 0;
}

这是我的 Mac 的输出（Mac OS X 10.10.1，GCC 4.9.1 — 为 SO 添加的额外换行符）：

$ ./ttt
UTC: {tm_sec = 54, tm_min = 59, tm_hour = 10, tm_mday = 20, tm_mon = 08, tm_year = 114,
tm_wday = 6, tm_yday = 262, tm_isdst = 0, tm_gmtoff = 00000, tm_zone = "UTC"}
Local: {tm_sec = 54, tm_min = 59, tm_hour = 03, tm_mday = 20, tm_mon = 08, tm_year = 114,
tm_wday = 6, tm_yday = 262, tm_isdst = 1, tm_gmtoff = -25200, tm_zone = "PDT"}
$ TZ="Europe/London" ./ttt
UTC: {tm_sec = 54, tm_min = 59, tm_hour = 10, tm_mday = 20, tm_mon = 08, tm_year = 114,
tm_wday = 6, tm_yday = 262, tm_isdst = 0, tm_gmtoff = 00000, tm_zone = "UTC"}
Local: {tm_sec = 54, tm_min = 59, tm_hour = 11, tm_mday = 20, tm_mon = 08, tm_year = 114,
tm_wday = 6, tm_yday = 262, tm_isdst = 1, tm_gmtoff = 03600, tm_zone = "BST"}
$

机器的默认时区设置实际上是 America/Los_Angeles（尽管 TZ 环境变量在第一次调用时正式取消设置）。如您所见，UTC 值是一致的，但本地时间值具有本地时区缩写。

在没有相反信息的情况下，我不得不得出结论，尽管您不这么认为，但您实际上是在调用 localtime() 而不是 gmtime()。您可以尝试通过在函数调用前添加 #undef gmtime 来确保您调用了 gmtime() 函数。如果 gmtime() 函数仍然产生错误的时间信息，那么您机器上的 gmtime() 就坏了。然而，这似乎极不可能。

在没有夏令时转换的情况下将纪元时间转换为人类日期时间字符串

Convert epoch time to human date time string without daylight saving conversion

c

time

epoch