Peter Zijlstra <pet...@infradead.org> writes: > On Tue, Apr 11, 2017 at 07:21:05AM +0530, Madhavan Srinivasan wrote: >> From: Sukadev Bhattiprolu <suka...@linux.vnet.ibm.com> >> >> perf_mem_data_src is an union that is initialized via the ->val field >> and accessed via the bitmap fields. For this to work on big endian >> platforms (Which is broken now), we also need a big-endian represenation >> of perf_mem_data_src. i.e, in a big endian system, if user request >> PERF_SAMPLE_DATA_SRC (perf report -d), will get the default value from >> perf_sample_data_init(), which is PERF_MEM_NA. Value for PERF_MEM_NA >> is constructed using shifts: >> >> /* TLB access */ >> #define PERF_MEM_TLB_NA 0x01 /* not available */ >> ... >> #define PERF_MEM_TLB_SHIFT 26 >> >> #define PERF_MEM_S(a, s) \ >> (((__u64)PERF_MEM_##a##_##s) << PERF_MEM_##a##_SHIFT) >> >> #define PERF_MEM_NA (PERF_MEM_S(OP, NA) |\ >> PERF_MEM_S(LVL, NA) |\ >> PERF_MEM_S(SNOOP, NA) |\ >> PERF_MEM_S(LOCK, NA) |\ >> PERF_MEM_S(TLB, NA)) >> >> Which works out as: >> >> ((0x01 << 0) | (0x01 << 5) | (0x01 << 19) | (0x01 << 24) | (0x01 << 26)) >> >> Which means the PERF_MEM_NA value comes out of the kernel as 0x5080021 >> in CPU endian. >> >> But then in the perf tool, the code uses the bitfields to inspect the >> value, and currently the bitfields are defined using little endian >> ordering. >> >> So eg. in perf_mem__tlb_scnprintf() we see: >> data_src->val = 0x5080021 >> op = 0x0 >> lvl = 0x0 >> snoop = 0x0 >> lock = 0x0 >> dtlb = 0x0 >> rsvd = 0x5080021 >> >> Patch does a minimal fix of adding big endian definition of the bitfields >> to match the values that are already exported by the kernel on big endian. >> And it makes no change on little endian. > > I think it is important to note that there are no current big-endian > users. So 'fixing' this will not break anybody and will ensure future > users (next patch) will work correctly.
Sure I'll fold in something along those lines. > Aside from that amendment, > > Acked-by: Peter Zijlstra (Intel) <pet...@infradead.org> Thanks. cheers