From: Jiong Wang jiong.wang@netronome.com
[ Upstream commit 69e168ebdcfcb87ce7252d4857d570f99996fa27 ]
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 Signed-off-by: Sasha Levin sashal@kernel.org --- drivers/net/ethernet/netronome/nfp/bpf/jit.c | 13 ++++++++++++- 1 file changed, 12 insertions(+), 1 deletion(-)
diff --git a/drivers/net/ethernet/netronome/nfp/bpf/jit.c b/drivers/net/ethernet/netronome/nfp/bpf/jit.c index 4e18d95e548f1..c3ce0fb47a0f0 100644 --- a/drivers/net/ethernet/netronome/nfp/bpf/jit.c +++ b/drivers/net/ethernet/netronome/nfp/bpf/jit.c @@ -326,7 +326,18 @@ __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 |