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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: 03f5781b ("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>
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