apply: remove epoch date from regex
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>
René Scharfe committed
Aug 25, 2017 at 21:06 UTC
0db3dc75f30239aa3b36071c7b9ff21c16ba449a
1 file changed
+5
-7
apply.c
+5
-7
@@ -812,9 +812,7 @@ static int has_epoch_timestamp(const char *nameline)
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* 1970-01-01, and the seconds part must be "00".
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*/
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const char stamp_regexp[] =
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- "^(1969-12-31|1970-01-01)"
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- " "
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- "[0-2][0-9]:[0-5][0-9]:00(\\.0+)?"
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+ "^[0-2][0-9]:([0-5][0-9]):00(\\.0+)?"
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" "
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"([-+][0-2][0-9]:?[0-5][0-9])\n";
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const char *timestamp = NULL, *cp, *colon;
@@ -834,9 +832,9 @@ static int has_epoch_timestamp(const char *nameline)
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* YYYY-MM-DD hh:mm:ss must be from either 1969-12-31
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* (west of GMT) or 1970-01-01 (east of GMT)
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*/
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- if (starts_with(timestamp, "1969-12-31"))
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+ if (skip_prefix(timestamp, "1969-12-31 ", ×tamp))
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epoch_hour = 24;
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- else if (starts_with(timestamp, "1970-01-01"))
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+ else if (skip_prefix(timestamp, "1970-01-01 ", ×tamp))
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epoch_hour = 0;
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else
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return 0;
@@ -858,8 +856,8 @@ static int has_epoch_timestamp(const char *nameline)
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return 0;
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}
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- hour = strtol(timestamp + 11, NULL, 10);
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- minute = strtol(timestamp + 14, NULL, 10);
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+ hour = strtol(timestamp, NULL, 10);
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+ minute = strtol(timestamp + m[1].rm_so, NULL, 10);
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zoneoffset = strtol(timestamp + m[3].rm_so + 1, (char **) &colon, 10);
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if (*colon == ':')