c++ - 如何在 C++ 中将 struct tm 转换为 time_t

Question

给定的函数是用于处理日期和时间的类的一部分。我解析的文件需要将给定的字符串数据转换为 time_t，但 mktime 不起作用。为什么？

 struct tm DateTimeUtils::makeTime(string arrTime)//accepts in format"2315"means 11.15 pm
{
    struct tm neww;
    string hour = arrTime.substr(0,2);
    int hour_int = stoi(hour);
    neww.tm_hour=hour_int;//when this is directly printed generates correct value


    string minute = arrTime.substr(2,2);
    int minute_int = stoi(minute);
    neww.tm_min=(minute_int);//when this is directly printed generates correct value

    time_t t1 = mktime(&neww);//only returns -1
    cout<<t1;

    return neww;

}

Answer 1

来自mktime(3) man page :

time_t ... represents the number of seconds elapsed since the Epoch, 1970-01-01 00:00:00 +0000 (UTC).

然后你有 struct tm 的字段，尤其是这个:

tm_year

The number of years since 1900.

所以基本上，如果 tm_year 设置为 0 并且我们正确地进行数学运算，我们将得到一个 70 年差，这需要以秒表示，这可能太大了。

您可以通过将 struct tm 值初始化为 Epoch 并将其用作基本引用来解决此问题:

 struct tm DateTimeUtils::makeTime(string arrTime)//accepts in format"2315"means 11.15 pm
{
    time_t tmp = { 0 };
    struct tm neww = *localtime(&tmp);
    string hour = arrTime.substr(0,2);
    int hour_int = stoi(hour);
    neww.tm_hour=hour_int;//when this is directly printed generates correct value


    string minute = arrTime.substr(2,2);
    int minute_int = stoi(minute);
    neww.tm_min=(minute_int);//when this is directly printed generates correct value

    time_t t1 = mktime(&neww);//only returns -1
    cout<<t1;

    return neww;
}