Searched refs:num_allocs (Results 1 – 4 of 4) sorted by relevance
36 int num_allocs = kMaxNumAllocs / 16; in main() local 37 for (int i = 0; num_allocs <= kMaxNumAllocs; i++, num_allocs *= 2) { in main() 38 fprintf(stderr, "[%d] allocating %d times\n", i, num_allocs); in main() 40 for (int j = 0; j < num_allocs; j++) { in main() 41 if ((j % (num_allocs / 8)) == 0) { in main() 54 if (zero_results != num_allocs) in main() 56 num_allocs - zero_results); in main() 57 for (int j = 0; j < num_allocs; j++) { in main()
10 result.num_allocs = 42; in Stop()
111 memory_iterations ? static_cast<double>(memory_result->num_allocs) / in CreateRunReport()
384 : num_allocs(0), in Result()390 int64_t num_allocs; member