use time

-- fibers — the writeonce hybrid demonstrated (BEAM's shape):
--   cooperative MECHANICS (no signals, no interrupts) with
--   reduction-budget EVICTION (loop back-edges pay one reduction; at
--   zero the fiber re-queues whether it likes it or not), actors as the
--   spawn surface (one message at a time, ownership-moving sends), and
--   blocking builtins that PARK on the shard's io_uring plane instead
--   of holding the thread.
--
-- Part 1 is deterministic by construction (budget accounting, no time):
-- its output is byte-exact. Part 2 shows a sleeping fiber not blocking
-- anyone; its ordering is asserted loosely by the acceptance gate
-- because it depends on real time.

class Tick {
  n: Int
}

-- Part 1: a counting actor; main and the actor interleave one message
-- per main-yield under the default budget.
class Counter {
  label: Text
  total: Int
  fn receive(msg: Tick) {
    self.total = self.total + msg.n;
    print("${self.label} +${msg.n} = ${self.total}");
  }
}

-- Part 2: an actor whose receive PARKS mid-message. The park releases
-- the shard: main keeps ticking while this fiber sleeps.
class Sleeper {
  pad: Int
  fn receive(msg: Tick) {
    print("sleeper: down for ${msg.n}ms");
    time.sleep(msg.n);
    print("sleeper: up");
  }
}

fn spin(rounds: Int) {
  -- burn reductions so the scheduler's eviction gets a chance: each
  -- back-edge pays one reduction; the default budget is 4000
  let i = 0;
  while i < rounds {
    i = i + 1;
  }
}

fn main() -> Int {
  -- ---- part 1: deterministic budget interleave ----
  let c: actor Tick = spawn Counter { label: "count", total: 0 };
  send(c, Tick { n: 1 });
  send(c, Tick { n: 2 });
  send(c, Tick { n: 3 });
  -- three yields deliver exactly three messages, one per turn
  spin(5000);
  spin(5000);
  spin(5000);
  print("part1 done");

  -- ---- part 2: a parked fiber blocks nobody ----
  let s: actor Tick = spawn Sleeper { pad: 0 };
  send(s, Tick { n: 150 });
  let t = 0;
  while t < 8 {
    time.sleep(25);
    print("main tick ${t}");
    t = t + 1;
  }
  print("part2 done");
  return 0;
}