Hi Zhouyi, On Wed, Apr 26, 2023 at 09:31:17AM +0800, Zhouyi Zhou wrote: [..] > Joel makes the learning process easier for me, indeed! I know that feeling being a learner myself ;-) > One question I have tried very hard to understand, but still confused. > for now, I know > r13 is fixed, but r1 is not, why "r9,40(r1)"'s 40(r1) can be assumed > to be equal to 3192(r10). First you have to I guess read up a bit about stack canaries. Google for "gcc stack protector" and "gcc stack canaries", and the look for basics of "buffer overflow attacks". That'll explain the concept of stack guards etc (Sorry if this is too obvious but I did not know how much you knew about it already). 40(r1) is where the canary was stored. In the beginning of the function, you have this: c000000000226d58: 78 0c 2d e9 ld r9,3192(r13) c000000000226d5c: 28 00 21 f9 std r9,40(r1) r1 is your stack pointer. 3192(r13) is the canary value. 40(r1) is where the canary is stored for later comparison. r1 should not change through out the function I believe, because otherwise you don't know where the stack frame is, right? Later you have this stuff before the function returns which gcc presumably did due to optimization. That mr means move register and is where the caching of r13 to r10 happens that Boqun pointed. c000000000226eb4: 78 6b aa 7d mr r10,r13 [...] and then the canary comparison happens: c000000000226ed8: 28 00 21 e9 ld r9,40(r1) c000000000226edc: 78 0c 4a e9 ld r10,3192(r10) c000000000226ee0: 79 52 29 7d xor. r9,r9,r10 c000000000226ee4: 00 00 40 39 li r10,0 c000000000226ee8: b8 03 82 40 bne c0000000002272a0 <srcu_gp_start_if_needed+0x5a0> So looks like for this to blow up, the preemption/migration has to happen precisely between the mr doing the caching, and the xor doing the comparison, since that's when the r10 will be stale. thanks, - Joel
