optimize two-way strstr and memmem bad character shift

first, the condition (mem && k < p) is redundant, because mem being
nonzero implies the needle is periodic with period exactly p, in which
case any byte that appears in the needle must appear in the last p
bytes of the needle, bounding the shift (k) by p.

second, the whole point of replacing the shift k by mem (=l-p) is to
prevent shifting by less than mem when discarding the memory on shift,
in which case linear time could not be guaranteed. but as written, the
check also replaced shifts greater than mem by mem, reducing the
benefit of the shift. there is no possible benefit to this reduction of
the shift; since mem is being cleared, the full shift is valid and
more optimal. so only replace the shift by mem when it would be less
than mem.

Author: richfelker

src/string/memmem.c

@@ -100,7 +100,7 @@ static char *twoway_memmem(const unsigned char *h, const unsigned char *z, const
 		if (BITOP(byteset, h[l-1], &)) {
 			k = l-shift[h[l-1]];
 			if (k) {
-				if (mem && k < p) k = l-p;
+				if (k < mem) k = mem;
 				h += k;
 				mem = 0;
 				continue;

src/string/strstr.c

@@ -109,7 +109,7 @@ static char *twoway_strstr(const unsigned char *h, const unsigned char *n)
 		if (BITOP(byteset, h[l-1], &)) {
 			k = l-shift[h[l-1]];
 			if (k) {
-				if (mem && k < p) k = l-p;
+				if (k < mem) k = mem;
 				h += k;
 				mem = 0;
 				continue;