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builtin-apply.c: restructure "offset" matching

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This restructures code to find matching location with offset
in find_offset() function, so that there is need for only one
call site of match_fragment() function.  There still isn't a
change in the logic of the program.

Signed-off-by: Junio C Hamano <gitster@pobox.com>
This commit is contained in:
Junio C Hamano 2008-01-19 02:16:16 -08:00
parent c89fb6b19a
commit fcb77bc57b
1 changed files with 26 additions and 25 deletions
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51
builtin-apply.c
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@ -1452,8 +1452,8 @@ static int find_offset(const char *buf, unsigned long size,
const char *fragment, unsigned long fragsize,
int line, int *lines)
{
int i;
unsigned long start, backwards, forwards;
int i, no_more_backwards, no_more_forwards;
unsigned long start, backwards, forwards, try;
if (fragsize > size)
return -1;
@ -1471,32 +1471,44 @@ static int find_offset(const char *buf, unsigned long size,
}
}
/* Exact line number? */
if (match_fragment(buf, size, start, fragment, fragsize))
return start;
/*
* There's probably some smart way to do this, but I'll leave
* that to the smart and beautiful people. I'm simple and stupid.
*/
backwards = start;
forwards = start;
for (i = 0; ; i++) {
unsigned long try;
int n;
try = start;
for (i = 0; ; i++) {
no_more_backwards = !backwards;
no_more_forwards = (forwards + fragsize > size);
if (match_fragment(buf, size, try, fragment, fragsize)) {
int shift = ((i+1) >> 1);
if (i & 1)
shift = -shift;
*lines = shift;
return try;
}
again:
if (no_more_backwards && no_more_forwards)
break;
/* "backward" */
if (i & 1) {
if (!backwards) {
if (forwards + fragsize > size)
break;
continue;
if (no_more_backwards) {
i++;
goto again;
}
do {
--backwards;
} while (backwards && buf[backwards-1] != '\n');
try = backwards;
} else {
if (no_more_forwards) {
i++;
goto again;
}
while (forwards + fragsize <= size) {
if (buf[forwards++] == '\n')
break;
@ -1504,18 +1516,7 @@ static int find_offset(const char *buf, unsigned long size,
try = forwards;
}
if (!match_fragment(buf, size, try, fragment, fragsize))
continue;
n = (i >> 1)+1;
if (i & 1)
n = -n;
*lines = n;
return try;
}
/*
* We should start searching forward and backward.
*/
return -1;
}