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apply: remove epoch date from regex

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We check the date of epoch timestamp candidates already with
starts_with().  Move beyond that part using skip_prefix() instead of
checking it again using a regular expression.  Also group the minutes
part, so that we can access them using a substring match instead of
using a magic number.

Signed-off-by: Rene Scharfe <l.s.r@web.de>
Signed-off-by: Junio C Hamano <gitster@pobox.com>
This commit is contained in:
René Scharfe 2017-08-25 21:06:28 +02:00 committed by Junio C Hamano
parent e4905019df
commit 0db3dc75f3
1 changed files with 5 additions and 7 deletions
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12
apply.c
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@ -812,9 +812,7 @@ static int has_epoch_timestamp(const char *nameline)
* 1970-01-01, and the seconds part must be "00".
*/
const char stamp_regexp[] =
"^(1969-12-31|1970-01-01)"
" "
"[0-2][0-9]:[0-5][0-9]:00(\\.0+)?"
"^[0-2][0-9]:([0-5][0-9]):00(\\.0+)?"
" "
"([-+][0-2][0-9]:?[0-5][0-9])\n";
const char *timestamp = NULL, *cp, *colon;
@ -834,9 +832,9 @@ static int has_epoch_timestamp(const char *nameline)
* YYYY-MM-DD hh:mm:ss must be from either 1969-12-31
* (west of GMT) or 1970-01-01 (east of GMT)
*/
if (starts_with(timestamp, "1969-12-31"))
if (skip_prefix(timestamp, "1969-12-31 ", &timestamp))
epoch_hour = 24;
else if (starts_with(timestamp, "1970-01-01"))
else if (skip_prefix(timestamp, "1970-01-01 ", &timestamp))
epoch_hour = 0;
else
return 0;
@ -858,8 +856,8 @@ static int has_epoch_timestamp(const char *nameline)
return 0;
}
hour = strtol(timestamp + 11, NULL, 10);
minute = strtol(timestamp + 14, NULL, 10);
hour = strtol(timestamp, NULL, 10);
minute = strtol(timestamp + m[1].rm_so, NULL, 10);
zoneoffset = strtol(timestamp + m[3].rm_so + 1, (char **) &colon, 10);
if (*colon == ':')