trailer: be stricter in parsing separators

Currently, a line is interpreted to be a trailer line if it contains a
separator. Make parsing stricter by requiring the text on the left of
the separator, if not the empty string, to be of the "<token><optional
whitespace>" form.

Signed-off-by: Jonathan Tan <jonathantanmy@google.com>
Signed-off-by: Junio C Hamano <gitster@pobox.com>
This commit is contained in:
Jonathan Tan 2016-11-02 10:29:16 -07:00 committed by Junio C Hamano
parent be5a750939
commit e4319562bc

View File

@ -563,15 +563,32 @@ static int token_matches_item(const char *tok, struct arg_item *item, int tok_le
}
/*
* Return the location of the first separator in line, or -1 if there is no
* separator.
* If the given line is of the form
* "<token><optional whitespace><separator>..." or "<separator>...", return the
* location of the separator. Otherwise, return -1. The optional whitespace
* is allowed there primarily to allow things like "Bug #43" where <token> is
* "Bug" and <separator> is "#".
*
* The separator-starts-line case (in which this function returns 0) is
* distinguished from the non-well-formed-line case (in which this function
* returns -1) because some callers of this function need such a distinction.
*/
static int find_separator(const char *line, const char *separators)
{
int loc = strcspn(line, separators);
if (!line[loc])
return -1;
return loc;
int whitespace_found = 0;
const char *c;
for (c = line; *c; c++) {
if (strchr(separators, *c))
return c - line;
if (!whitespace_found && (isalnum(*c) || *c == '-'))
continue;
if (c != line && (*c == ' ' || *c == '\t')) {
whitespace_found = 1;
continue;
}
break;
}
return -1;
}
/*