bpf, x32: Fix bug with ALU64 {LSH, RSH, ARSH} BPF_X shift by 0

bpf, x32: Fix bug with ALU64 {LSH, RSH, ARSH} BPF_X shift by 0

The current x32 BPF JIT for shift operations is not correct when the
shift amount in a register is 0. The expected behavior is a no-op, whereas
the current implementation changes bits in the destination register.

The following example demonstrates the bug. The expected result of this
program is 1, but the current JITed code returns 2.

  r0 = 1
  r1 = 1
  r2 = 0
  r1 <<= r2
  if r1 == 1 goto end
  r0 = 2
end:
  exit

The bug is caused by an incorrect assumption by the JIT that a shift by
32 clear the register. On x32 however, shifts use the lower 5 bits of
the source, making a shift by 32 equivalent to a shift by 0.

This patch fixes the bug using double-precision shifts, which also
simplifies the code.

Fixes: 03f5781be2c7 ("bpf, x86_32: add eBPF JIT compiler for ia32")
Co-developed-by: Xi Wang <xi.wang@gmail.com>
Signed-off-by: Xi Wang <xi.wang@gmail.com>
Signed-off-by: Luke Nelson <luke.r.nels@gmail.com>
Signed-off-by: Daniel Borkmann <daniel@iogearbox.net>