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

borkmann · davem330 · commit 6205b9cf200d · 2016-02-21T22:07:11.000-05:00
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: Daniel Borkmann &lt;daniel@iogearbox.net&gt;
Acked-by: Alexei Starovoitov &lt;ast@kernel.org&gt;
Signed-off-by: David S. Miller &lt;davem@davemloft.net&gt;
diff --git a/net/core/filter.c b/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;
 
