ASN1_TIME 到 time_t 的转换 [英] ASN1_TIME to time_t conversion

问题描述

如何将 ASN1_TIME 转换为 time_t 格式?我想将 X509_get_notAfter() 的返回值转换为秒.

How can I convert ASN1_TIME to time_t format? I wanted to convert the return value of X509_get_notAfter() to seconds.

推荐答案

时间在内部存储为字符串，格式为 YYmmddHHMMSS 或 YYYYmmddHHMMSS.

Times are stored as a string internally, on the format YYmmddHHMMSS or YYYYmmddHHMMSS.

在字符串的末尾有几分之一秒和时区的空间，但现在让我们忽略它，并有一些(未经测试的)代码.

At the end of the string there is room for fractions of seconds and timezone, but let's ignore that for now, and have some (untested) code.

注意:另请参阅下面 Bryan Olson 的回答，其中讨论了由于 i++ 导致的未定义行为.另请参阅 Seak 的答案，它消除了未定义的行为.

Note: also see Bryan Olson's answer below, which discusses the undefined behavior due to the i++'s. Also see Seak's answer which removes the undefined behavior.

static time_t ASN1_GetTimeT(ASN1_TIME* time)
{
    struct tm t;
    const char* str = (const char*) time->data;
    size_t i = 0;

    memset(&t, 0, sizeof(t));

    if (time->type == V_ASN1_UTCTIME) /* two digit year */
    {
        t.tm_year = (str[i++] - '0') * 10 + (str[++i] - '0');
        if (t.tm_year < 70)
        t.tm_year += 100;
    }
    else if (time->type == V_ASN1_GENERALIZEDTIME) /* four digit year */
    {
        t.tm_year = (str[i++] - '0') * 1000 + (str[++i] - '0') * 100 + (str[++i] - '0') * 10 + (str[++i] - '0');
        t.tm_year -= 1900;
    }
    t.tm_mon = ((str[i++] - '0') * 10 + (str[++i] - '0')) - 1; // -1 since January is 0 not 1.
    t.tm_mday = (str[i++] - '0') * 10 + (str[++i] - '0');
    t.tm_hour = (str[i++] - '0') * 10 + (str[++i] - '0');
    t.tm_min  = (str[i++] - '0') * 10 + (str[++i] - '0');
    t.tm_sec  = (str[i++] - '0') * 10 + (str[++i] - '0');

    /* Note: we did not adjust the time based on time zone information */
    return mktime(&t);
}

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