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: Daniel Borkmann <daniel@iogearbox.net>
Acked-by: Alexei Starovoitov <ast@kernel.org>
Signed-off-by: David S. Miller <davem@davemloft.net>

diff --git a/net/core/filter.c b/net/core/filter.c
index 8a0b8c3eb189fc2d95f123f2b2005547a1f0a938..a3aba15a802544ee87d768637c54054db9f914c2 100644 (file)
--- a/net/core/filter.c
+++ b/net/core/filter.c
@@ -530,12 +530,14 @@ do_pass:
                        *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;