bpf: do not use reciprocal divide

At first Jakub Zawadzki noticed that some divisions by reciprocal_divide
were not correct. (off by one in some cases)
http://www.wireshark.org/~darkjames/reciprocal-buggy.c

He could also show this with BPF:
http://www.wireshark.org/~darkjames/set-and-dump-filter-k-bug.c

The reciprocal divide in linux kernel is not generic enough,
lets remove its use in BPF, as it is not worth the pain with
current cpus.

Signed-off-by: Eric Dumazet <edumazet@google.com>
Reported-by: Jakub Zawadzki <darkjames-ws@darkjames.pl>
Cc: Mircea Gherzan <mgherzan@gmail.com>
Cc: Daniel Borkmann <dxchgb@gmail.com>
Cc: Hannes Frederic Sowa <hannes@stressinduktion.org>
Cc: Matt Evans <matt@ozlabs.org>
Cc: Martin Schwidefsky <schwidefsky@de.ibm.com>
Cc: Heiko Carstens <heiko.carstens@de.ibm.com>
Cc: David S. Miller <davem@davemloft.net>
Signed-off-by: David S. Miller <davem@davemloft.net>

arch/s390/net/bpf_jit_comp.c

@@ -371,11 +371,13 @@ static int bpf_jit_insn(struct bpf_jit *jit, struct sock_filter *filter,
 		/* dr %r4,%r12 */
 		EMIT2(0x1d4c);
 		break;
-	case BPF_S_ALU_DIV_K: /* A = reciprocal_divide(A, K) */
-		/* m %r4,<d(K)>(%r13) */
-		EMIT4_DISP(0x5c40d000, EMIT_CONST(K));
-		/* lr %r5,%r4 */
-		EMIT2(0x1854);
+	case BPF_S_ALU_DIV_K: /* A /= K */
+		if (K == 1)
+			break;
+		/* lhi %r4,0 */
+		EMIT4(0xa7480000);
+		/* d %r4,<d(K)>(%r13) */
+		EMIT4_DISP(0x5d40d000, EMIT_CONST(K));
 		break;
 	case BPF_S_ALU_MOD_X: /* A %= X */
 		jit->seen |= SEEN_XREG | SEEN_RET0;
@@ -391,6 +393,11 @@ static int bpf_jit_insn(struct bpf_jit *jit, struct sock_filter *filter,
 		EMIT2(0x1854);
 		break;
 	case BPF_S_ALU_MOD_K: /* A %= K */
+		if (K == 1) {
+			/* lhi %r5,0 */
+			EMIT4(0xa7580000);
+			break;
+		}
 		/* lhi %r4,0 */
 		EMIT4(0xa7480000);
 		/* d %r4,<d(K)>(%r13) */