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If first_sg and last_sg wraps around in the scatterlist ring, then we need to account for that in the shift as well. E.g. crafting such msgs where this is the case leads to a hang as shift becomes negative. E.g. consider the following scenario: first_sg := 14 |=> shift := -12 msg->sg_start := 10 last_sg := 3 | msg->sg_end := 5 round 1: i := 15, move_from := 3, sg[15] := sg[ 3] round 2: i := 0, move_from := -12, sg[ 0] := sg[-12] round 3: i := 1, move_from := -11, sg[ 1] := sg[-11] round 4: i := 2, move_from := -10, sg[ 2] := sg[-10] [...] round 13: i := 11, move_from := -1, sg[ 2] := sg[ -1] round 14: i := 12, move_from := 0, sg[ 2] := sg[ 0] round 15: i := 13, move_from := 1, sg[ 2] := sg[ 1] round 16: i := 14, move_from := 2, sg[ 2] := sg[ 2] round 17: i := 15, move_from := 3, sg[ 2] := sg[ 3] [...] This means we will loop forever and never hit the msg->sg_end condition to break out of the loop. When we see that the ring wraps around, then the shift should be MAX_SKB_FRAGS - first_sg + last_sg - 1. Meaning, the remainder slots from the tail of the ring and the head until last_sg combined. Fixes: 015632bb ("bpf: sk_msg program helper bpf_sk_msg_pull_data") Signed-off-by:Daniel Borkmann <daniel@iogearbox.net> Acked-by:
John Fastabend <john.fastabend@gmail.com> Signed-off-by:
Alexei Starovoitov <ast@kernel.org>
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