Question

I’m using spawn_link but doesn’t understand its behavior. Consider the following code:

-module(test).
-export([try_spawn_link/0]).

try_spawn_link() ->
  spawn(fun() ->
    io:format("parent: ~p~n", [Parent = self()]),
    Client = spawn_link(fun() ->
      io:format("child: ~p~n", [self()]),
      spawn_link_loop(Parent)
    end),
    spawn_link_loop(Client)
  end).

spawn_link_loop(Peer) ->
  receive
    quit ->
      exit(normal);
    Any ->
      io:format("~p receives ~p~n", [self(), Any])
  end,
  spawn_link_loop(Peer).

From the Erlang documentation, a link is created between the calling process and the new process, atomically. However, I tested as follows and didn’t notice the effect of the link.

1> test:try_spawn_link().
parent: <0.34.0>
<0.34.0>
child: <0.35.0>
2> is_process_alive(pid(0,34,0)).
true
3> is_process_alive(pid(0,35,0)).
true
4> pid(0,35,0) ! quit.
quit
5> is_process_alive(pid(0,35,0)).
false
6> is_process_alive(pid(0,34,0)).
true

1> test:try_spawn_link().
parent: <0.34.0>
<0.34.0>
child: <0.35.0>
2> is_process_alive(pid(0,34,0)).
true
3> is_process_alive(pid(0,35,0)).
true
4> pid(0,34,0) ! quit.
quit
5> is_process_alive(pid(0,35,0)).
true
6> is_process_alive(pid(0,34,0)).
false

In my understanding, if one peer of the link exits, the other peer exits (or is notified to exit). But the results seem different from my understanding.

EDIT: thanks to the answers of legoscia and Pascal.

Answer 1

As described in the “Error handling” section of the Processes chapter of the Erlang reference manual, a linked process exiting causes its linked processes to exit only if the exit reason is not normal. That’s why OTP extensively uses the shutdown exit reason.