[signa11]

> Erlang does a great job of protecting data from concurrency problems, but you're still vulnerable to mistakes like deadlock.

without any shared data, and an asynchronous-message-passing style concurrency (i.e sharing memory by communicating, as opposed to communicating by sharing memory) can you please elucidate how we might get into deadlocks ? thanks !

[macintux]

OTP offers two basic messaging options: async or synchronous. The latter is implemented by forcing the caller to wait for the response.

If process A sends a synchronous message to B, and B while processing it sends a different synchronous message either directly to A or to another process that eventually calls back to A, you end up in a deadlocked state.

[signa11]

> OTP offers two basic messaging options: async or synchronous.

with synchronous messaging it is quite easy to see how deadlocks can happen, without trying too hard. for async messaging, which is what i was referring to, it is quite hard (and or convoluted) to get into that state...

[Jtsummers]

receive is a blocking action in erlang so something like:

  loop(State) ->
    receive
      msg -> loop(State)
    end.

Will block waiting for `msg` to show up. So if you have two processes that are interacting with eachother and aren't careful in constructing their communication you could end up with something (contrived):

  loop(Pid) ->
    receive
      msg -> Pid ! msg
    end,
    loop(Pid).

If you have two processes A and B that end up in this same loop but referencing each other, they'd deadlock. Substituting A and B for Pid you'd end up with something like:

  loop(B) ->    % Process A
    receive
      msg -> B ! msg
    end,
    loop(B).
  loop(A) ->    % Process B
    receive
      msg -> A ! msg
    end,
    loop(A).

Both waiting for `msg` but never sending it to their partner process.

[signa11]

> Both waiting for `msg` but never sending it to their partner process.

but then you can 'deadlock' for a single pid as well right ? where you wait for a message which no one is sending...

[Jtsummers]

A single process could wait forever if no message ever arrives, yes. But it's not a deadlock since deadlock requires at least two processes that are waiting on each other in some fashion.

And responding to your lower down comment: Yes, if a third party could send `msg` to either of these it'd break the deadlock. In my example (and the cases I've caused this myself) there was no other process around to do that. But this is where, with experience, you learn to design things better and also use `after` clauses in the receive expression. This will cause them to time out and do something. Which may be to terminate and let a supervisor restart the whole thing or some other behavior.

[sudhirj]

That wouldn't count as a deadlock. A deadlock implies that it's possible to make a logical determination that the system is permanently stuck, and that state cannot possibly change without external intervention.

i.e. a deadlock can only be determined when you look at the system as a whole from the outside and determine that it's permanently stuck.

[signa11]

> That wouldn't count as a deadlock. A deadlock implies that it's possible to make a logical determination that the system is permanently stuck, and that state cannot possibly change without external intervention. i.e. a deadlock can only be determined when you look at the system as a whole from the outside and determine that it's permanently stuck.

yes, i know.

as i have pointed out elsewhere, with synchronous invocations, it is easy to get into a loop (or deadlock) i.e. pid-a -> pid-b -> pid-c -> pid-a. for the async case, which is what was elucidated by gp, it seems to me that the example is 'deadlocking' only because no one is sending messages expected by the other party...