ebpf: remove self-assignment in interpreter's tail call

ARG1 = BPF_R1 as it stands, evaluates to regs[BPF_REG_1] = regs[BPF_REG_1]
and thus has no effect. Add a comment instead, explaining what happens and
why it's okay to just remove it. Since from user space side, a tail call is
invoked as a pseudo helper function via bpf_tail_call_proto, the verifier
checks the arguments just like with any other helper function and makes
sure that the first argument (regs[BPF_REG_1])'s type is ARG_PTR_TO_CTX.
Signed-off-by: Daniel Borkmann <daniel@iogearbox.net>
Acked-by: Alexei Starovoitov <ast@plumgrid.com>
Signed-off-by: David S. Miller <davem@davemloft.net>

kernel/bpf/core.c

@@ -453,7 +453,11 @@ static unsigned int __bpf_prog_run(void *ctx, const struct bpf_insn *insn)
 		if (unlikely(!prog))
 			goto out;
 
-		ARG1 = BPF_R1;
+		/* ARG1 at this point is guaranteed to point to CTX from
+		 * the verifier side due to the fact that the tail call is
+		 * handeled like a helper, that is, bpf_tail_call_proto,
+		 * where arg1_type is ARG_PTR_TO_CTX.
+		 */
 		insn = prog->insnsi;
 		goto select_insn;
 out: