Private Channels

Problem

Multiple clients communicate with one server. How does each client receive only its own replies without building separate channels?

Solution: tag replies with client id

The server tags each reply with the originating client id. Clients filter the shared outbox by their own id.

sequenceDiagram
  participant Bus as shared-bus
  participant S as server-c
  participant A as client-a filter
  participant B as client-b filter

  Bus->>S: {id="a", value=5}
  Bus->>S: {id="b", value=3}
  Bus->>S: {id="a", value=10}
  S-->>A: {id="a", result=10}
  S-->>B: {id="b", result=6}
  S-->>A: {id="a", result=20}

server-beh = msg: reply { id = msg.id; result = msg.value * 2; };
server-c = actor server-beh;

shared-bus =
  st
    { id = "a"; value = 5; }
    { id = "b"; value = 3; }
    { id = "a"; value = 10; };

server = server-c { inbox = shared-bus; };

client-a = server.outbox.filter (r: r.id == "a");
client-b = server.outbox.filter (r: r.id == "b");

client-a.toList
# [ { id = "a"; result = 10; } { id = "a"; result = 20; } ]
client-b.toList
# [ { id = "b"; result = 6; } ]

Each client sees only its slice. The server needs no routing logic — it just echoes the id back.

With Either results

Combine the tagged bus with reply.right / reply.left for per-client error separation:

compute =
  msg:
  if msg.divisor == 0 then { left = "div-by-zero"; }
  else { right = msg.dividend / msg.divisor; };

server-c = actor (msg: reply { id = msg.id; result = compute msg; });

shared-bus =
  st
    { id = "a"; dividend = 10; divisor = 2; }
    { id = "b"; dividend = 6;  divisor = 0; }
    { id = "a"; dividend = 9;  divisor = 3; };

server = server-c { inbox = shared-bus; };

a-replies = server.outbox.filter (r: r.id == "a");
b-replies = server.outbox.filter (r: r.id == "b");

Each client gets { id, result } where result is { right = n } or { left = msg }.

Alternative: isolated sessions via when-c

If clients want full per-client state (not just reply filtering), use when-c to give each client its own actor instance:

flowchart TB
  shared["shared bus\n[{to=alice,cmd=inc},{to=bob,cmd=inc},{to=alice,cmd=get}]"]
  wca["when-c to==alice → cmd"]
  wcb["when-c to==bob → cmd"]
  a["alice\ncounter-c\nstate: 2"]
  b["bob\ncounter-c\nstate: 1"]
  shared --> wca --> a
  shared --> wcb --> b

shared =
  st
    { to = "alice"; cmd = "inc"; }
    { to = "bob";   cmd = "inc"; }
    { to = "alice"; cmd = "get"; };

alice = counter-c { inbox = when-c (m: m.to == "alice") shared (m: m.cmd); };
bob   = counter-c { inbox = when-c (m: m.to == "bob")   shared (m: m.cmd); };

Alice and Bob each get their own counter state. Messages for one never affect the other.

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