> Am 17.07.2019 um 18:25 schrieb Y Song <ys114321@xxxxxxxxx>: > > On Wed, Jul 17, 2019 at 3:36 AM Ilya Leoshkevich <iii@xxxxxxxxxxxxx> wrote: >> >> >> Here is a better one: len=0x11223344 and we would like to do >> ((u8 *)&len)[3]. >> >> len is represented as `11 22 33 44` in memory, so the desired result is >> 0x44. It can be obtained by doing (*(u32 *)&len) & 0xff, but today the >> verifier does ((*(u32 *)&len) >> 24) & 0xff instead. > > What you described above for the memory layout all makes sense. > The root cause is for big endian, we should do *((u8 *)&len + 3). > This is exactly what macros in test_pkt_md_access.c tries to do. > > if __BYTE_ORDER__ == __ORDER_LITTLE_ENDIAN__ > #define TEST_FIELD(TYPE, FIELD, MASK) \ > { \ > TYPE tmp = *(volatile TYPE *)&skb->FIELD; \ > if (tmp != ((*(volatile __u32 *)&skb->FIELD) & MASK)) \ > return TC_ACT_SHOT; \ > } > #else > #define TEST_FIELD_OFFSET(a, b) ((sizeof(a) - sizeof(b)) / sizeof(b)) > #define TEST_FIELD(TYPE, FIELD, MASK) \ > { \ > TYPE tmp = *((volatile TYPE *)&skb->FIELD + \ > TEST_FIELD_OFFSET(skb->FIELD, TYPE)); \ > if (tmp != ((*(volatile __u32 *)&skb->FIELD) & MASK)) \ > return TC_ACT_SHOT; \ > } > #endif > > Could you check whether your __BYTE_ORDER__ is set > correctly or not for this case? You may need to tweak Makefile > if you are doing cross compilation, I am not sure how as I > did not have environment. I’m building natively on s390. Here is the (formatted) preprocessed C code for the first condition: { __u8 tmp = *((volatile __u8 *)&skb->len + ((sizeof(skb->len) - sizeof(__u8)) / sizeof(__u8))); if (tmp != ((*(volatile __u32 *)&skb->len) & 0xFF)) return 2; }; So I believe the endianness is chosen correctly. Here is the clang-generated BPF bytecode for the first condition: # llvm-objdump -d test_pkt_md_access.o 0000000000000000 process: 0: 71 21 00 03 00 00 00 00 r2 = *(u8 *)(r1 + 3) 1: 61 31 00 00 00 00 00 00 r3 = *(u32 *)(r1 + 0) 2: 57 30 00 00 00 00 00 ff r3 &= 255 3: 5d 23 00 1d 00 00 00 00 if r2 != r3 goto +29 <LBB0_10> This also looks good to me. Finally, here is the verifier-generated BPF bytecode: # bpftool prog dump xlated id 14 ; TEST_FIELD(__u8, len, 0xFF); 0: (61) r2 = *(u32 *)(r1 +104) 1: (bc) w2 = w2 2: (74) w2 >>= 24 3: (bc) w2 = w2 4: (54) w2 &= 255 5: (bc) w2 = w2 Here we can see the shift that I'm referring to. I believe we should translate *(u8 *)(r1 + 3) in this case without this shift on big-endian machines. Best regards, Ilya