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