# another test of else with multiple processes
# note: processes synchronize so execution is deterministic;
# output is also deterministic.
resource a()
# note:  talk about being sensitive...
#        removing hoho's second parameter makes infinite loop disappear
#
   op bozo(int t)
   op hoho(int t, int u)
   op blocker() returns bool b
   op f(int x)
   op go()
   op done()
   # control[t] is 0 if else present for hoho during try t
   # otherwise it is value to send to f
const int R = 2
      int  control[1:R] =
	  ( 29,
	     0)
      int  output[1:R] = ([R] -1)

      for [ t = 1 to R ] {
	  write("****Try",t)
	  send f(9999)
	  if (control[t] != 0) { send f(control[t])  }
	  call bozo(t)
	  call hoho(t,t)
	  receive done()
      }
      for [ k = 1 to R ] {
	    write(output[k])
      }
      for [ k = 1 to R st control[k] != output[k] ] {
	    write("oops",k); stop(1)
      }

   proc bozo(t)  {
      write("bozo",t)
      reply
      # don't try this at home boys and girls
      in f(x) and blocker() ->
      ni
   }

   proc hoho(t,u)  {
      reply
send go()
      if (control[t] == 0) {
	  in f(x) and x == -98765 /* always false */ -> output[t] = x
	  [] else -> output[t] = 0
	  ni
      } else {
	  in f(x) -> output[t] = x
	  ni
      }
      send done()
   }
   # the idea is to block bozo so that the hoho's get put on
   # the new process list for f.
   proc blocker() returns b  {
      b = true
receive go()
   }
end