YARB: Yet Another Ring Benchmark, Part I
March 21, 2009In Joe Armstrong’s excellent Erlang book he has an exercise to:
Write a ring benchmark. Create N processes in a ring. Send a message round the
ring M times so that a total or N*M messages get sent. Time how long it takes
for different values of N and M. Write a similar program in some other language
you are familar with. Compare the results. Write a blog and public the results
on the Internet!
So far I’ve done the Erlang section of this. It turns out to be pretty straight forward in Erlang, not surprisingly. If it was difficult it wouldn’t be a great example now would it.
So lets see how it’s done.
-module(ring).
-compile(export_all).
-import(lists, [foreach/2, map/2, reverse/1]).
main([A,B]) ->
N = list_to_integer(A),
M = list_to_integer(B),
io:format("~w: ~w processes with ~w messages\n", [self(), N, M]),
This section simply reads in the command-line arguments and uses list_to_integer() to convert them into numbers.
[First|_] = Pids = for(1, N, fun() -> spawn(fun() -> loop() end) end),
setupLoop(Pids),
Next we spawn the required number of threads and setup the loop, so that each thread knows the PID of the next thread in the loop.
sendMsgs(M, First),
% Cleanup!!
map(fun(P) -> P ! {exit} end, Pids).
SengMsgs does the work of sending messages around the loop. It starts with sending a {relay} message to the first thread in the loop and waits for a reply from the last thread in the loop. When it receives a message from the last thread, it decrements the message count and sends another {relay} message. Continuing until the message count reaches 0.
sendMsgs(0, First) ->
First ! {relay},
receive
{relay} ->
io:format("~w: Finished ~w loops.~n", [self(), 0])
end;
sendMsgs(N, First) ->
First ! {relay},
receive
{relay} ->
sendMsgs(N-1, First)
end.
To setup the loop, we take the first 2 PIDs sending the second PID to the first as a message indicating that it’s the next thread in the loop. When we get down to the last PID, the next thread is the main thread.
The second language to compare against, needs to be something with a similar threading model with support for something similar to message passing. I’m looking at doing it with either Haskell or Lisp but I haven’t worked out how to do the message passing just yet. It’ll have to wait for PartII.
Hope this helps someone, and please ask questions if my explaination isn’t clear.
Full source code follows, simply copy the code and place into a file called ring.erl. Compile using this command erlc ring.erl and run the resulting program like so
erl -noshell -run ring main 1 2 -run init stop
-module(ring).
-compile(export_all).
-import(lists, [foreach/2, map/2, reverse/1]).
%%% Write a ring benchmark. Create N processes in a ring. Send a
%%% message round the ring M times so that a total of N*M messages
%%% get sent. Time how long this takes for different values of N and M.
%%% Write a similar program in some other language you are familar with.
%%% Compare the results. Write a blog, and publish the results on the Internet!
%%% Compile: "erlc ring.erl" Run: "erl -noshell -run ring main 1 2 -run init stop"
main([A,B]) ->
N = list_to_integer(A),
M = list_to_integer(B),
io:format("~w: ~w processes with ~w messages\n", [self(), N, M]),
% Spawn 'N' processes and link them together.
[First|_] = Pids = for(1, N, fun() -> spawn(fun() -> loop() end) end),
setupLoop(Pids),
% Start sending messages.
sendMsgs(M, First),
% Cleanup!!
map(fun(P) -> P ! {exit} end, Pids).
setupLoop([P]) ->
P ! {next, self()};
setupLoop([P1,P2|Rest]) ->
P1 ! {next, P2},
setupLoop([P2|Rest]).
sendMsgs(0, First) ->
First ! {relay},
receive
{relay} ->
io:format("~w: Finished ~w loops.~n", [self(), 0])
end;
sendMsgs(N, First) ->
First ! {relay},
receive
{relay} ->
sendMsgs(N-1, First)
end.
loop() ->
io:format("~w: Starting node~w~n", [self(), self()]),
receive
{next, Pid} ->
loop(Pid);
{exit} ->
io:format("~w: Exiting now! ~n", [self()])
end.
loop(Pid) ->
receive
{relay} ->
Pid ! {relay},
loop(Pid);
{exit} ->
io:format("~w: Exiting now!~n", [self()])
end.
for(N, N, F) ->
[F()];
for(I, N, F) ->
[F() | for (I+1, N, F)].