bpf: don't emit mov A,A on return

While debugging with bpf_jit_disasm I noticed emissions of 'mov %eax,%eax',
and found that this comes from BPF_RET | BPF_A translations from classic
BPF. Emitting this is unnecessary as BPF_REG_A is mapped into BPF_REG_0
already, therefore only emit a mov when immediates are used as return value.
Signed-off-by: NDaniel Borkmann <daniel@iogearbox.net>
Acked-by: NAlexei Starovoitov <ast@kernel.org>
Signed-off-by: NDavid S. Miller <davem@davemloft.net>

net/core/filter.c

@@ -530,12 +530,14 @@ static int bpf_convert_filter(struct sock_filter *prog, int len,
 			*insn = BPF_MOV64_REG(BPF_REG_A, BPF_REG_TMP);
 			break;
 
-		/* RET_K, RET_A are remaped into 2 insns. */
+		/* RET_K is remaped into 2 insns. RET_A case doesn't need an
+		 * extra mov as BPF_REG_0 is already mapped into BPF_REG_A.
+		 */
 		case BPF_RET | BPF_A:
 		case BPF_RET | BPF_K:
-			*insn++ = BPF_MOV32_RAW(BPF_RVAL(fp->code) == BPF_K ?
-						BPF_K : BPF_X, BPF_REG_0,
-						BPF_REG_A, fp->k);
+			if (BPF_RVAL(fp->code) == BPF_K)
+				*insn++ = BPF_MOV32_RAW(BPF_K, BPF_REG_0,
+							0, fp->k);
 			*insn = BPF_EXIT_INSN();
 			break;