When spawning a remote process like this, does the ‘work’ (1+2 in this case) get evaluated on the local node?
I’d like the ‘work’ to be evaluated on the remote node, and when I replace 1+2 with Node.self(), it appears that Node.self()is evaluated locally, and that the remote node is sending the already computed value back to the local node:
iex(dev@127.0.0.1)7> Node.spawn(:"suv@127.0.0.1", Kernel, :send, [caller, {:response, Node.self}])
#PID<13803.123.0>
iex(dev@127.0.0.1)8> flush
{:response, :"dev@127.0.0.1}
:ok
I’m a relative Elixir noob, so not quite grasping how to make the remote node do the work ![]()
Thanks for your help


















