ref-filter: simplify lstrip_ref_components() memory handling

We're walking forward in the string, skipping path components from
left-to-right. So when we've stripped as much as we want, the pointer we
have is a complete NUL-terminated string and we can just return it
(after duplicating it, of course). So there is no need for a temporary
allocated string.

But we do make an extra temporary copy due to f0062d3 (ref-filter:
free item->value and item->value->s, 2018-10-18). This is probably from
cargo-culting the technique used in rstrip_ref_components(), which
_does_ need a separate string (since it is stripping from the end and
ties off the temporary string with a NUL).

Let's drop the extra allocation. This is slightly more efficient, but
more importantly makes the code much simpler.

Signed-off-by: Jeff King <peff@peff.net>
Signed-off-by: Junio C Hamano <gitster@pobox.com>

Author: peff

ref-filter.c

@@ -2196,23 +2196,18 @@ static int normalize_component_count(const char *refname, int len)
 static const char *lstrip_ref_components(const char *refname, int len)
 {
 	int remaining = normalize_component_count(refname, len);
-	const char *start = xstrdup(refname);
-	const char *to_free = start;
 
 	while (remaining > 0) {
-		switch (*start++) {
+		switch (*refname++) {
 		case '\0':
-			free((char *)to_free);
 			return xstrdup("");
 		case '/':
 			remaining--;
 			break;
 		}
 	}
 
-	start = xstrdup(start);
-	free((char *)to_free);
-	return start;
+	return xstrdup(refname);
 }
 
 static const char *rstrip_ref_components(const char *refname, int len)