将字符串从__DATE__转换为time_t [英] Convert string from __DATE__ into a time_t

问题描述

我正在尝试将从 __ DATE __ 宏生成的字符串转换为 time_t 。我不需要一个完整的日期/时间解析器，只处理 __ DATE __ 宏格式的格式是很棒的。

I'm trying to convert the string produced from the __DATE__ macro into a time_t. I don't need a full-blown date/time parser, something that only handles the format of the __DATE__ macro would be great.

一个预处理器的方法会很漂亮，但一个函数也可以工作。如果相关，我使用MSVC。

A preprocessor method would be nifty, but a function would work just as well. If it's relevant, I'm using MSVC.

推荐答案

编辑：更正的函数应该是这样的：

the corrected function should look something like this:

time_t cvt_TIME(char const *time) { 
    char s_month[5];
    int month, day, year;
    struct tm t = {0};
    static const char month_names[] = "JanFebMarAprMayJunJulAugSepOctNovDec";

    sscanf(time, "%s %d %d", s_month, &day, &year);

    month = (strstr(month_names, s_month)-month_names)/3;

    t.tm_mon = month;
    t.tm_mday = day;
    t.tm_year = year - 1900;
    t.tm_isdst = -1;

    return mktime(&t);
}

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