Openssl
代码中提供了函数
ASN1_TIME_set
，把
time_t
格式的时间转化为
ASN1_TIME
函数原型为
ASN1_TIME *ASN1_TIME_set(ASN1_TIME *s, time_t t)
但是没有找到类似
ASN1_TIME_get
函数，把
ASN1_TIME
转化为
time_t
从网上找个代码，实现了
ASN1_TIME_get
，测试结果正确。
[win2003 VC6
下测试通过
]
代码：
int
mypint(
const
char
** s,
int
n,
int
min,
int
max,
int
* e
)
{
int
retval = 0;
while
(n) {
if
(**s < '0' || **s > '9') { *e = 1;
return
0; }
retval *= 10;
retval += **s - '0';
--n; ++(*s);
}
if
(retval < min || retval > max) *e = 1;
return
retval;
}
time_t
ASN1_TIME_get (
ASN1_TIME
* a,
int
*err
)
{
char
days[2][12] =
{
{ 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 },
{ 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 }
};
int
dummy;
const
char
*s;
int
generalized;
struct
tm t;
int
i, year, isleap, offset;
time_t
retval;
if
(err == NULL) err = &dummy;
if
(a->type == V_ASN1_GENERALIZEDTIME) {
generalized = 1;
}
else
if
(a->type == V_ASN1_UTCTIME) {
generalized = 0;
}
else
{
*err = 1;
return
0;
}
s = a->data;
// Data should be always null terminated
if
(s == NULL || s[a->length] != '\0') {
*err = 1;
return
0;
}
*err = 0;
if
(generalized) {
t.tm_year = mypint(&s, 4, 0, 9999, err) - 1900;
}
else
{
t.tm_year = mypint(&s, 2, 0, 99, err);
if
(t.tm_year < 50) t.tm_year += 100;
}
t.tm_mon = mypint(&s, 2, 1, 12, err) - 1;
t.tm_mday = mypint(&s, 2, 1, 31, err);
// NOTE: It's not yet clear, if this implementation is 100% correct
// for GeneralizedTime... but at least misinterpretation is
// impossible --- we just throw an exception
t.tm_hour = mypint(&s, 2, 0, 23, err);
t.tm_min = mypint(&s, 2, 0, 59, err);
if
(*s >= '0' && *s <= '9') {
t.tm_sec = mypint(&s, 2, 0, 59, err);
}
else
{
t.tm_sec = 0;
}
if
(*err)
return
0;
// Format violation
if
(generalized) {
// skip fractional seconds if any
while
(*s == '.' || *s == ',' || (*s >= '0' && *s <= '9')) ++s;
// special treatment for local time
if
(*s == 0) {
t.tm_isdst = -1;
retval = mktime(&t);
// Local time is easy :)
if
(retval == (time_t)-1) {
*err = 2;
retval = 0;
}
return
retval;
}
}
if
(*s == 'Z') {
offset = 0;
++s;
}
else
if
(*s == '-' || *s == '+') {
i = (*s++ == '-');
offset = mypint(&s, 2, 0, 12, err);
offset *= 60;
offset += mypint(&s, 2, 0, 59, err);
if
(*err)
return
0;
// Format violation
if
(i) offset = -offset;
}
else
{
*err = 1;
return
0;
}
if
(*s) {
*err = 1;
return
0;
}
// And here comes the hard part --- there's no standard function to
// convert struct tm containing UTC time into time_t without
// messing global timezone settings (breaks multithreading and may
// cause other problems) and thus we have to do this "by hand"
//
// NOTE: Overflow check does not detect too big overflows, but is
// sufficient thanks to the fact that year numbers are limited to four
// digit non-negative values.
retval = t.tm_sec;
retval += (t.tm_min - offset) * 60;
retval += t.tm_hour * 3600;
retval += (t.tm_mday - 1) * 86400;
year = t.tm_year + 1900;
if
(
sizeof
(time_t) == 4) {
// This is just to avoid too big overflows being undetected, finer
// overflow detection is done below.
if
(year < 1900 || year > 2040) *err = 2;
}
// FIXME: Does POSIX really say, that all years divisible by 4 are
// leap years (for consistency)??? Fortunately, this problem does
// not exist for 32-bit time_t and we should'nt be worried about
// this until the year of 2100 :)
isleap = ((year % 4 == 0) && (year % 100 != 0)) || (year % 400 == 0);
for
(i = t.tm_mon - 1; i >= 0; --i) retval += days[isleap][i] * 86400;
retval += (year - 1970) * 31536000;
if
(year < 1970) {
retval -= ((1970 - year + 2) / 4) * 86400;
if
(
sizeof
(time_t) > 4) {
for
(i = 1900; i >= year; i -= 100) {
if
(i % 400 == 0)
continue
;
retval += 86400;
}
}
if
(retval >= 0) *err = 2;
}
else
{
retval += ((year - 1970 + 1) / 4) * 86400;
if
(
sizeof
(time_t) > 4) {
for
(i = 2100; i < year; i += 100) {
// The following condition is the reason to
// start with 2100 instead of 2000
if
(i % 400 == 0)
continue
;
retval -= 86400;
}
}
if
(retval < 0) *err = 2;
}
if
(*err) retval = 0;
return
retval;
}