bench: Assert that division by zero is unreachable #9547

because of the ++count above I don't believe this can be 1 here.

@gmaxwell But there is a --count on the line below :-)

yes, my read it can't be 1 or 0, it must be >1 and then the decrements reduces it to 1. If it is possible to be 0 at the division then I believe there is a bug elsewhere (in addition).

If count would be guaranteed to be different from zero, then the check on line 50 (if (count == 0) {) wouldn't be needed, right? :-)

If the branch if (count == 0) { is taken then the relevant parts of the remaining code can be reduced to something like:

count = 0; // implicit
++count;
--count;
double average = (now-beginTime)/count;

Assuming I'm reading it correctly :-)

Let me know if I'm missing something obvious – this is one of my first contributions so please bear with me :-)

Looks like if maxElapsed is > 0 then the logic here is redundant for that code path. count=0 means now and beginTime are both 0.

It might be better to fix the issue in the output at the bottom of the function like so:

     // Output results
-    double average = (now-beginTime)/count;
-    int64_t averageCycles = (nowCycles-beginCycles)/count;
+    double average = (count > 0) ? (now-beginTime)/count : 0.0;
+    int64_t averageCycles = (count > 0) ? (nowCycles-beginCycles)/count : 0;
     std::cout << std::fixed << std::setprecision(15) << name << "," << count << "," << minTime << "," << maxTime << "," << average << ","
               << minCycles << "," << maxCycles << "," << averageCycles << "\n";

I would say that this reads ok in isolation as just dumb output code and avoids having to understand the branching in function body above.