ASN1_TIME转换为time_t [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

时间在内部以字符串形式存储，格式为YYmmddHHMMSS或YYYYmmddHHMMSS.

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

该字符串的末尾有几分之一秒和时区的空间，但是现在让我们忽略它，并保留一些(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 :另请参见下面的布莱恩·奥尔森(Bryan Olson)的答案，其中讨论了由于i++导致的不确定行为.另请参阅Seak的答案，该答案删除了未定义的行为.

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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