Question: PCIe Eventgroups

[HelgeBrands]

Hi,
I have a question to readout PCI-E data. In an earlier version there were two functions used (programPCIeCounters and programPCIeMissCounters) now there are programPCIeEventGroup/getPCIeCounterData. For me it is not completly clear where and how to setup these enventgroups. In the example code these Platform stuff is now used (getPlatform). This is not in the main library only in the examples.
Can you give me some hints how to generate these groups.

Thanks for your help
Best regards
Helge Brands

[opcm]

the event groups are setup here: https://github.com/opcm/pcm/blob/0a178ec2573821e3c08beb6bd76808ebb912bc76/pcm-pcie.h#L368

hope this helps..

[berry-sh]

Hello,

I want to get PCIRdCur (no. of reads to LLC) in my .cpp file, so what I did was using programPCIeEventGroup() and passed to it a vector of events that I want to monitor, more specifically PCIRdCur_miss (uint_64 of {0x00043c33}) and PCIRdCur_hit (uint_64 of {0x00043c37}).

First of all, is that the correct way to determine which events to monitor? And given that my CPU's microarchitecture is CLX, can I program more than 1 event to monitor? The reason I am asking is that I found inside CPUCounters.cpp the following sentence: "//JKT through СLX generations allow programming only one required event at a time"

Then, I called getPCIeCounterData(socket_no, ctr). The thing is that I am not sure what to specify as ctr. My understanding is that ctr ranges from 0 to 3 (otherwise, it gives segmentation fault).

The question is if the 0-3 range of the ctr variable corresponds to (0)PCIRdCur(RFO/CRd/DRd), and (1)ItoM, and (2)PRd, and (3)WiL, respectively. The reason I guessed so is because in pcm-pcie.h, it loops from ctr = 0 to ctr = eventGroup.size(), so I assumed ctr is used to represent these events. Is this right tho?

If that is the case, why does getPCIeCounterData function still displays values when ctr = 1 (monitoring ItoM) while I have already programmed the event group to only monitor PCIRdCur?

Thanks for your help in advance.