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|
(* TEST
flags = "-g"
* skip
reason = "port stat-mem-prof : https://github.com/ocaml/ocaml/pull/8634"
*)
open Gc.Memprof
let rec allocate_list accu = function
| 0 -> accu
| n -> allocate_list (n::accu) (n-1)
let[@inline never] allocate_lists len cnt =
for j = 0 to cnt-1 do
ignore (allocate_list [] len)
done
let check_distrib len cnt rate =
Printf.printf "check_distrib %d %d %f\n%!" len cnt rate;
let smp = ref 0 in
start ~callstack_size:10 ~sampling_rate:rate
{ null_tracker with
alloc_major = (fun _ -> assert false);
alloc_minor = (fun info ->
assert (info.size = 2);
assert (info.n_samples > 0);
assert (info.source = Normal);
smp := !smp + info.n_samples;
None);
};
allocate_lists len cnt;
stop ();
(* The probability distribution of the number of samples follows a
binomial distribution of parameters tot_alloc and rate. Given
that tot_alloc*rate and tot_alloc*(1-rate) are large (i.e., >
100), this distribution is approximately equal to a normal
distribution. We compute a 1e-8 confidence interval for !smp
using quantiles of the normal distribution, and check that we are
in this confidence interval. *)
let tot_alloc = cnt*len*3 in
assert (float tot_alloc *. rate > 100. &&
float tot_alloc *. (1. -. rate) > 100.);
let mean = float tot_alloc *. rate in
let stddev = sqrt (float tot_alloc *. rate *. (1. -. rate)) in
(* This assertion has probability to fail close to 1e-8. *)
assert (abs_float (mean -. float !smp) <= stddev *. 5.7)
let () =
check_distrib 10 1000000 0.01;
check_distrib 1000000 10 0.00001;
check_distrib 1000000 10 0.0001;
check_distrib 1000000 10 0.001;
check_distrib 1000000 10 0.01;
check_distrib 100000 10 0.1;
check_distrib 100000 10 0.9
let () =
Printf.printf "OK !\n"
|
