nfp: bpf: fix static check error through tightening shift amount adjustment

NFP shift instruction has something special. If shift direction is left
then shift amount of 1 to 31 is specified as 32 minus the amount to shift.

But no need to do this for indirect shift which has shift amount be 0. Even
after we do this subtraction, shift amount 0 will be turned into 32 which
will eventually be encoded the same as 0 because only low 5 bits are
encoded, but shift amount be 32 will fail the FIELD_PREP check done later
on shift mask (0x1f), due to 32 is out of mask range. Such error has been
observed when compiling nfp/bpf/jit.c using gcc 8.3 + O3.

This issue has started when indirect shift support added after which the
incoming shift amount to __emit_shf could be 0, therefore it is at that
time shift amount adjustment inside __emit_shf should have been tightened.

Fixes: 991f5b3651f6 ("nfp: bpf: support logic indirect shifts (BPF_[L|R]SH | BPF_X)")
Reported-by: Oleksandr Natalenko <oleksandr@natalenko.name>
Reported-by: Pablo Cascón <pablo.cascon@netronome.com
Reviewed-by: Quentin Monnet <quentin.monnet@netronome.com>
Reviewed-by: Jakub Kicinski <jakub.kicinski@netronome.com>
Signed-off-by: Jiong Wang <jiong.wang@netronome.com>
Signed-off-by: Alexei Starovoitov <ast@kernel.org>

diff --git a/drivers/net/ethernet/netronome/nfp/bpf/jit.c b/drivers/net/ethernet/netronome/nfp/bpf/jit.c
index f272247d170830b60ad95b617a0446a50cb0961d..d4bf0e694541feb6e1d8b8031328f3305a85a893 100644 (file)
--- a/drivers/net/ethernet/netronome/nfp/bpf/jit.c
+++ b/drivers/net/ethernet/netronome/nfp/bpf/jit.c
@@ -328,7 +328,18 @@ __emit_shf(struct nfp_prog *nfp_prog, u16 dst, enum alu_dst_ab dst_ab,
                return;
        }
 
-       if (sc == SHF_SC_L_SHF)
+       /* NFP shift instruction has something special. If shift direction is
+        * left then shift amount of 1 to 31 is specified as 32 minus the amount
+        * to shift.
+        *
+        * But no need to do this for indirect shift which has shift amount be
+        * 0. Even after we do this subtraction, shift amount 0 will be turned
+        * into 32 which will eventually be encoded the same as 0 because only
+        * low 5 bits are encoded, but shift amount be 32 will fail the
+        * FIELD_PREP check done later on shift mask (0x1f), due to 32 is out of
+        * mask range.
+        */
+       if (sc == SHF_SC_L_SHF && shift)
                shift = 32 - shift;
 
        insn = OP_SHF_BASE |