On Fri, Apr 18, 2025 at 09:53:59AM -0500, Andrew Hamilton wrote:
> Support dates outside of 1901..2038.
>
> Fixes: https://savannah.gnu.org/bugs/?63894
> Fixes: https://savannah.gnu.org/bugs/?66301
>
> Signed-off-by: Vladimir Serbinenko <phco...@gmail.com>
> Signed-off-by: Andrew Hamilton <adham...@gmail.com>
> ---
>  grub-core/lib/datetime.c | 48 ++++++++++++++++++++++++++++++++--------
>  include/grub/datetime.h  | 15 ++++++-------
>  2 files changed, 46 insertions(+), 17 deletions(-)
>
> diff --git a/grub-core/lib/datetime.c b/grub-core/lib/datetime.c
> index 8f0922fb0..4e68eabc6 100644
> --- a/grub-core/lib/datetime.c
> +++ b/grub-core/lib/datetime.c
> @@ -64,6 +64,10 @@ grub_get_weekday_name (struct grub_datetime *datetime)
>  #define SECPERDAY (24*SECPERHOUR)
>  #define DAYSPERYEAR 365
>  #define DAYSPER4YEARS (4*DAYSPERYEAR+1)
> +/* 24 leap years in 100 years */
> +#define DAYSPER100YEARS (100*DAYSPERYEAR+24)
> +/* 97 leap years in 400 years */
> +#define DAYSPER400YEARS (400*DAYSPERYEAR+97)
>
>
>  void
> @@ -76,7 +80,7 @@ grub_unixtime2datetime (grub_int64_t nix, struct 
> grub_datetime *datetime)
>    /* Convenience: let's have 3 consecutive non-bissextile years
>       at the beginning of the counting date. So count from 1901. */
>    int days_epoch;
> -  /* Number of days since 1st Januar, 1901.  */
> +  /* Number of days since 1st January, 1 (proleptic).  */
>    unsigned days;
>    /* Seconds into current day.  */
>    unsigned secs_in_day;
> @@ -92,12 +96,39 @@ grub_unixtime2datetime (grub_int64_t nix, struct 
> grub_datetime *datetime)
>      days_epoch = grub_divmod64 (nix, SECPERDAY, NULL);
>
>    secs_in_day = nix - days_epoch * SECPERDAY;
> -  days = days_epoch + 69 * DAYSPERYEAR + 17;
> +  /* 1970 is Unix Epoch. Adjust to a year 1 epoch:
> +     Leap year logic:
> +      - Years evenly divisible by 400 are leap years
> +      - Otherwise, if divisible by 100 are not leap years
> +      - Otherwise, if divisible by 4 are leap years
> +     There are four 400-year periods (1600 years worth of days with leap 
> days)
> +     There are three 100-year periods worth of leap days (3*24)
> +     There are 369 years in addition to the four 400 year periods
> +     There are 17 leap days in 69 years (beyond the three 100 year periods) 
> */

May I ask you to stick to comments coding style [1]?

> +  days = days_epoch + 369 * DAYSPERYEAR + 17 + 24 * 3 + 4 * DAYSPER400YEARS;
>
> -  datetime->year = 1901 + 4 * (days / DAYSPER4YEARS);
> +  datetime->year = 1 + 400 * (days / DAYSPER400YEARS);
> +  days %= DAYSPER400YEARS;
> +
> +  /* On 31st December of bissextile (leap) years 365 days from the beginning
> +     of the year elapsed but year isn't finished yet - every 400 years
> +     396 is 4 years less than 400 year leap cycle
> +     96 is 1 day less than number of leap days in 400 years */

Ditto... Here and below please...

> +  if (days / DAYSPER100YEARS == 4)
> +    {
> +      datetime->year += 396;
> +      days -= 396*DAYSPERYEAR + 96;
> +    }
> +  else
> +    {
> +      datetime->year += 100 * (days / DAYSPER100YEARS);
> +      days %= DAYSPER100YEARS;
> +    }
> +
> +  datetime->year += 4 * (days / DAYSPER4YEARS);
>    days %= DAYSPER4YEARS;
> -  /* On 31st December of bissextile years 365 days from the beginning
> -     of the year elapsed but year isn't finished yet */
> +  /* On 31st December of bissextile (leap) years 365 days from the beginning
> +     of the year elapsed but year isn't finished yet - every 4 years */
>    if (days / DAYSPERYEAR == 4)
>      {
>        datetime->year += 3;
> @@ -108,11 +139,10 @@ grub_unixtime2datetime (grub_int64_t nix, struct 
> grub_datetime *datetime)
>        datetime->year += days / DAYSPERYEAR;
>        days %= DAYSPERYEAR;
>      }
> +  int isbisextile = datetime->year % 4 == 0 && (datetime->year % 100 != 0 || 
> datetime->year % 400 == 0);

Please do not define variables in random places. I prefer if it is done
at the beginning of the function or block.

And I think this should be a bool set with ternary operator...

... and s/isbisextile/is_bisextile/

>    for (i = 0; i < 12
> -      && days >= (i==1 && datetime->year % 4 == 0
> -                   ? 29 : months[i]); i++)
> -    days -= (i==1 && datetime->year % 4 == 0
> -                         ? 29 : months[i]);
> +        && days >= (i==1 && isbisextile ? 29 : months[i]); i++)
> +    days -= (i==1 && isbisextile ? 29 : months[i]);

If you touch the code then please fix its coding style, e.g.:

  days -= (i == 1 && is_bisextile == true) ? 29 : months[i]

On the occasion it seems more readable...

>    datetime->month = i + 1;
>    datetime->day = 1 + days;
>    datetime->hour = (secs_in_day / SECPERHOUR);
> diff --git a/include/grub/datetime.h b/include/grub/datetime.h
> index bcec636f0..9289b0d00 100644
> --- a/include/grub/datetime.h
> +++ b/include/grub/datetime.h
> @@ -54,7 +54,7 @@ void grub_unixtime2datetime (grub_int64_t nix,
>  static inline int
>  grub_datetime2unixtime (const struct grub_datetime *datetime, grub_int64_t 
> *nix)
>  {
> -  grub_int32_t ret;
> +  grub_int64_t ret;
>    int y4, ay;
>    const grub_uint16_t monthssum[12]
>      = { 0,
> @@ -75,15 +75,11 @@ grub_datetime2unixtime (const struct grub_datetime 
> *datetime, grub_int64_t *nix)
>    const int SECPERHOUR = 60 * SECPERMIN;
>    const int SECPERDAY = 24 * SECPERHOUR;
>    const int SECPERYEAR = 365 * SECPERDAY;
> -  const int SECPER4YEARS = 4 * SECPERYEAR + SECPERDAY;
> +  const grub_int64_t SECPER4YEARS = 4 * SECPERYEAR + SECPERDAY;
>
> -  if (datetime->year > 2038 || datetime->year < 1901)
> -    return 0;
>    if (datetime->month > 12 || datetime->month < 1)
>      return 0;
>
> -  /* In the period of validity of unixtime all years divisible by 4
> -     are bissextile*/
>    /* Convenience: let's have 3 consecutive non-bissextile years
>       at the beginning of the epoch. So count from 1973 instead of 1970 */
>    ret = 3 * SECPERYEAR + SECPERDAY;
> @@ -94,13 +90,16 @@ grub_datetime2unixtime (const struct grub_datetime 
> *datetime, grub_int64_t *nix)
>    ret += y4 * SECPER4YEARS;
>    ret += ay * SECPERYEAR;
>
> +  ret -= ((datetime->year - 1) / 100 - (datetime->year - 1) / 400 - 15) * 
> SECPERDAY;

What does this line do? And where 15 comes from?

Daniel

_______________________________________________
Grub-devel mailing list
Grub-devel@gnu.org
https://lists.gnu.org/mailman/listinfo/grub-devel

Reply via email to