Following is the program in which i have tried to spawn 3 processes form a method called best. I want to receive response from all the processes and store them in a tuple but I am able to get only one response.
test() ->
receive
{From,N} -> From!{self(),N},
loop()
end.
best(N) ->
Aid=spawn(fun t:loop/0),
Aid ! {self(),N},
Bid=spawn(fun t:loop/0),
Bid ! {self(),N},
Cid=spawn(fun t:loop/0),
Cid ! {self(),N},
receive
{Pid,Response} ->{Response}
end.
Can someone please help me out with this probem
Your receive bloc, in the best/2 function exit as soon as it receives one message. If you launch this code in the shell, you can verify that the other message are still in the message queue with the function flush(). (The code you posted is missing the t:loop/0 function, I guess it will compute something based on N and return the answer via a message to the spawner)
To be able to receive more than one message, you must put the receive bloc in a "loop" that recursively calls itself until it got all answers. You will have to use a variable that allows the recursive loop to know when it is finished (number of answers expected, list of processes that should answer...) and collect the answers in a list variable for example.
-module(wy).
-compile(export_all).
loop() ->
Self = self(),
receive
{From, Ref, N} ->
From ! {Self, Ref, N * N}
end.
receive_result(Ref) ->
receive
{Pid, Ref, R} ->
io:format("process ~p: ~p~n", [Pid, R]),
receive_result(Ref)
after 10 ->
ok
end.
best() ->
APid = spawn(fun loop/0),
BPid = spawn(fun loop/0),
CPid = spawn(fun loop/0),
Self = self(),
Ref = make_ref(),
APid ! {Self, Ref, 2},
BPid ! {Self, Ref, 3},
CPid ! {Self, Ref, 4},
receive_result(Ref).
You can follow this small code. The result is:
9> wy:best().
process <0.77.0>: 4
process <0.78.0>: 9
process <0.79.0>: 16
ok
Related
experimenting with distributed erlang, here's what I have:
loop()->
receive {From, ping} ->
io:format("received ping from ~p~n", [From]),
From ! pong,
loop();
{From, Fun} when is_function(Fun) ->
io:format("executing function ~p received from ~p~n", [Fun, From]),
From ! Fun(),
loop()
end.
test_remote_node_can_execute_sent_clojure()->
Pid = spawn(trecias, fun([])-> loop() end),
Pid ! {self(), fun()-> erlang:nodes() end},
receive Result ->
Result = [node()]
after 300 ->
timeout
end.
getting: Can not start erlang:apply,[#Fun<tests.1.123107452>,[]] on trecias
node I execute the test on runs on the same machine as the node 'trecias'. Both nodes can load same code.
Any ideas what is amiss?
In the spawn call, you've specified the node name as trecias, but you need to specify the full node name including the hostname, e.g. trecias#localhost.
Also, the function you pass to spawn/2 must take zero arguments, but the one in the code above takes one argument (and crashes if that argument isn't the empty list). Write it as fun() -> loop() end instead.
When spawning an anonymous function on a remote node, you also need to make sure that the module is loaded on both nodes, with the same version. Otherwise you'll get a badfun error.
I just had a simple question and I can't seem to find an answer to it. Basically my question is if I have a function recording state (recursive) and I send multiple messages to it, will it keep going through the receive block until it no longer has messages in its "mailbox"?
state(Fridge) ->
receive
Pat1 ->
io:format("ok"),
state(S);
Pat2 ->
io:format("not ok"),
state(S)
end.
So if I'd send to this process 3 messages (Pat1, Pat2, Pat1) using "!" and its not able to go into its loop before receiving messages will it still print out the following?
1> "ok"
2> "not ok"
3> "ok"
Sorry if this isn't very clearly put, by simplifying the question it might make it hard to picture what I am asking.
Your question isn't clear but you seem to be asking whether the process will receive the three messages even if they're sent before the target process has called receive — if that's the question, the answer is yes. When you send a message to a process, it goes into that process's message queue until the process calls receive to remove it from the message queue and deal with it.
If you call erlang:process_info(Pid, messages) where Pid is the receiver's process id, you can see what messages are in its queue. You might try this from the Erlang shell.
As an extreme example of message queueing, under some heavy load conditions it can be a source of out-of-memory problems if a receiver can't keep up with a fast sender. Under these conditions, a receiver's message queue might grow without bound until the system runs out of memory.
Does this answer your question?
1> OUT = fun(X) -> io:format(">>> ~p~n", [X]) end.
#Fun<erl_eval.6.54118792>
2> F = fun X() -> receive foo -> OUT(foo), X(); bar -> OUT(bar), X() end end.
#Fun<erl_eval.44.54118792>
3> P = spawn(F).
<0.38.0>
4> [ P ! X || X <- [foo, bar, foo]].
>>> foo
[foo,bar,foo]
>>> bar
>>> foo
A message arrives in the mailbox and then receive patterns are applied to the message like the function clause or case statement. But unlike those, if none of them match, next message is processed and the previous one left in the message box untouched. Other receive clause starts always from beginning of message queue.
1> OUT = fun(X) -> io:format(">>> ~p~n", [X]) end.
#Fun<erl_eval.6.54118792>
2> F = fun() -> receive start -> (fun X() -> receive foo -> OUT(foo), X(); bar -> OUT(bar), X() end end)() end end.
#Fun<erl_eval.20.54118792>
3> P = spawn(F).
<0.38.0>
4> [ P ! X || X <- [foo, bar, foo]].
[foo,bar,foo]
5> P! start.
>>> foo
start
>>> bar
>>> foo
Note that foo, bar, foo is in the queue in the first receive but it is not processed. When start arrives (last in the queue) second receive starts process foo and bar messages.
-module(wy).
-compile(export_all).
state() ->
timer:sleep(2000),
io:format("~p~n", [erlang:process_info(self(), messages)]),
receive
pat1 ->
io:format("ok~n"),
state();
pat2 ->
io:format("not ok~n"),
state()
end.
main() ->
Pid = spawn(fun state/0),
[ Pid ! X || X <- [pat1, pat2, pat1]].
You can run this code, from the output you can understand everything.
**Note:**when you send message to a process, the sent messages will be stored in mail box of the process.
6> wy:main().
[pat1,pat2,pat1]
7> {messages,[pat1,pat2,pat1]}
7> ok
7> {messages,[pat2,pat1]}
7> not ok
7> {messages,[pat1]}
7> ok
7> {messages,[]}
7>
One comments for your code:
receive
Pat1 ->
io:format("ok"),
state(S);
Pat2 ->
io:format("not ok"),
state(S)
end.
the second clause will not match forever.
I want to write program which creates 2 processes.
First print number , send it to second process and so on.
However I have problem in start function.
-module(zad).
-export([start/0, one/0, two/0]).
one() ->
receive
{go1, N} ->
io:format("1 ~w~n", [N]),
two ! {go2, N+1},
one();
{go2, 50} ->
io:format("END ~w~n", [N+1]),
two ! finished
end.
two() ->
receive
finished ->
io:format("two finished~n", []);
{go2, N} ->
io:format("2 ~w~n", [N]),
one ! {go1, N+1 },
two()
end.
start() ->
register(two, spawn(zad, two, [])),
register(one, spawn(zad, one, [])),
one ! {go1, 0}.
here's error
** exception error: bad argument
in function register/2
called as register(two,<0.146.0>)
in call from zad:start/0 (zad.erl, line 29)
Does it mean that I cant register 2 processes in start? What is wrong with this code?
UPDATE
It is really weird. but after I changed
{go2, 50} -> io:format("END ~w~n", [N+1]),
to
{go2, 50} -> io:format("END ~w~n", [50]),
error disappeared.
It's really strange because problem is in one(), but error was in register(two)
You can register two processes in start/0. It means the values you are passing into register/2 are wrong. From the documentation(http://erlang.org/doc/man/erlang.html#register-2):
Failures:
badarg If PidOrPort is not an existing local process or port.
badarg If RegName is already in use.
badarg If the process or port is already registered (already has a name). badarg If RegName is the atom undefined.
Likely what has happened is the process died before register/2 was called (see the first failure in the list above).
From the Learn You Some Erlang for Great Good!
Another special case is when the timeout is at 0:
flush() ->
receive
_ -> flush()
after 0 ->
ok
end
.
When that happens, the Erlang VM will try and find a message that fits
one of the available patterns. In the case above, anything matches. As
long as there are messages, the flush/0 function will recursively call
itself until the mailbox is empty. Once this is done, the after 0 ->
ok part of the code is executed and the function returns.
I don't understand purpose of after 0. After reading above text I thought it was like after infinity (waiting forever) but I changed a little the flush function:
flush2() ->
receive
_ -> timer:sleep(1000), io:format("aa~n"), flush()
after 0 ->
okss
end
.
flush3() ->
receive
_ -> io:format("aa~n"), flush()
after 0 ->
okss
end
.
In the first function it waits 1 second and in the second function it doesn't wait.
In both cases it doesn't display a text (aa~n).
So it doesn't work as after infinity.
If block between the receive and the after are not executed then above 2 codes can be simplified to:
flush4() ->
okss
.
What I am missing?
ps. I am on the Erlang R16B03-1, and author of the book was, as fair I remember, was on the Erlang R13.
Every process has a 'mailbox' -- message queue. Messages can be fetched by receive. if there is no messages in the queue. after part specifies how much time 'receive will wait for them. So after 0 -- means process checking (by receive ) if any messages in the queue and if queue is empty immediately continue to next instructions.
It can be used for instance if we want periodically check if any messages here and to do something (hopefully helpful) if there is no messages.
Consider after 0 to be finally.
Consider the use of after 0 to process receives with a priority: http://learnyousomeerlang.com/more-on-multiprocessing#selective-receives
May this different look on things enlighten you.
You can play with the following shell command to understand the effect of the after command:
4> L = fun(G) ->
4> receive
4> stop -> ok;
4> M -> io:format("received ~p~n",[M]), G(G)
4> after 0 ->
4> io:format("no message~n")
4> end
4> end.
#Fun<erl_eval.6.80484245>
5> F = fun() -> timer:sleep(10000),
5> io:format("end of wait for messages, go to receive block~n"),
5> L(L)end.
#Fun<erl_eval.20.80484245>
6> spawn(F).
<0.46.0>
end of wait for messages, go to receive block
no message
7> P1 = spawn(F).
<0.52.0>
8> P1 ! hello.
hello
end of wait for messages, go to receive block
received hello
no message
9> P2 ! hello, P2 ! stop.
* 1: variable 'P2' is unbound
8> P2 = spawn(F).
<0.56.0>
9> P2 ! hello, P2 ! stop.
stop
end of wait for messages, go to receive block
received hello
10>
If you do not intend to use a nested receive, rather than using "after" part, I think a better approach is to use "Unexpected ->" variable to handle all unmatched messages.
I'm a beginner at Erlang and I've been working through "Learn You Some Erlang For Great Good!". I use a modified version of this example code where the critic has a parameter:
critic(Count) ->
receive
{From, {"Rage Against the Turing Machine", "Unit Testify"}} ->
From ! {self(), {"They are great!", Count}};
{From, {"System of a Downtime", "Memoize"}} ->
From ! {self(), {"They're not Johnny Crash but they're good.", Count}};
{From, {"Johnny Crash", "The Token Ring of Fire"}} ->
From ! {self(), {"Simply incredible.", Count}};
{From, {_Band, _Album}} ->
From ! {self(), {"They are terrible!", Count}}
end,
critic(Count).
Which is spawned like this:
restarter() ->
process_flag(trap_exit, true),
Pid = spawn_link(?MODULE, critic, [my_atom]),
register(critic, Pid),
receive
{'EXIT', Pid, normal} -> % not a crash
ok;
{'EXIT', Pid, shutdown} -> % manual termination, not a crash
ok;
{'EXIT', Pid, _} ->
restarter()
end.
The module is used like this:
1> c(linkmon).
{ok,linkmon}
2> Monitor = linkmon:start_critic().
<0.163.0>
3> linkmon:judge("Rage Against the Turing Machine", "Unit Testify").
{"They are great!",my_atom}
Now, when I change "my_atom" to a simple number (like 255) the monitor crashes:
1> c(linkmon).
{ok,linkmon}
2> Monitor = linkmon:start_critic().
=ERROR REPORT==== 14-Jul-2013::20:42:20 ===
Error in process <0.173.0> with exit value: {badarg,[{erlang,register,[critic,<0.174.0>] []},{linkmon,restarter,0,[{file,"linkmon.erl"},{line,16}]}]}
However, it does work when I send [1] (so the code is "spawn(....., [[255]]).")
Why can't I pass a single number? Is just skimming over the documentation of spawn/3 doesn't really tell me anything... except maybe that I missed something and a number is not an Erlang term. But then how do I pass a number?
The error message says that the call to register(critic, Pid) on line 16 crashes due to "badarg" even though the arguments look ok. This can happen if the process referred to by Pid is already dead (if it crashes immediately, e.g. if you pass the wrong number of args), or if you already have a process around using that name. Ensure that the length of the list in the spawn(Mod,Fun,[...]) matches the number of args to your critic() function, and call "whereis(critic)" in the shell to check if there's an old process blocking the name from being reused.