将字符串时间转换为 unix 时间，反之亦然

Question

我一直在尝试将字符串“1998-04-11”简单地转换为 UNIX 时间戳，根据在线转换器，它应该是 892245600。

但我总是得到不同的结果。

    struct tm tm;
    time_t ts;

    strptime("1998-04-11", "%Y-%m-%d", &tm);
    tm.tm_mon = tm.tm_mon -1;
    ts = mktime(&tm);

    printf("%d \n", (int)ts); //unix time-stamp
    printf("%s \n", ctime(&ts)); //human readable date

结果：

893502901
Sat Apr 25 13:15:01 1998

谁能告诉我我做错了什么？

Answer 1

在调用 strptime

之前将 tm 结构清零

memset(&tm, 0, sizeof(struct tm));

来自注释部分：http://man7.org/linux/man-pages/man3/strptime.3.html

In principle, this function does not initialize tm but stores only the values specified. This means that tm should be initialized before the call.

并且memset在同一页面的示例中如上所示使用。

Answer 2

这是未初始化内存的问题。

(gdb) p tm
 = {tm_sec = 1, tm_min = 0, tm_hour = 4196061, tm_mday = 0, tm_mon = -5984, tm_year = 32767, 
tm_wday = 0, tm_yday = 0, tm_isdst = 4195984, tm_gmtoff = 4195616, 
tm_zone = 0x7fffffffe980 "[=10=]1"}

如您在调试器中所见，struct tm 分配了随机内存。制作 time_zone 偏移量垃圾。

strptime 运行后：

(gdb) p tm
 = {tm_sec = 1, tm_min = 0, tm_hour = 4196061, tm_mday = 11, tm_mon = 3, tm_year = 98, 
tm_wday = 6, tm_yday = 100, tm_isdst = 4195984, tm_gmtoff = 4195616, 
tm_zone = 0x7fffffffe980 "[=11=]1"}

此外：

tm.tm_mon = tm.tm_mon -1;

不需要。更正后的代码：

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

int main(int argc, char **argv) {

struct tm tm;
time_t ts = 0;
memset(&tm, 0, sizeof(tm));

    strptime("1998-04-11", "%Y-%m-%d", &tm);
    ts = mktime(&tm);

    printf("%d \n", (int)ts); //unix time-stamp
    printf("%s \n", ctime(&ts)); //human readable date
}

输出：

892252800
Sat Apr 11 00:00:00 1998

将字符串时间转换为 unix 时间，反之亦然

Convert string time to unix time and vice versa

c++

unix-timestamp