问jheatbugs-2001-03-28中某些代码

sybbk

HeatbugModelSwarm中buildActions部分，3个try分别是做什么？查了下refbook-java-2.2，解释太简略，还是不懂，高手指点，谢谢！代码如下：

public Object buildActions () {
. |5 t# R" r4 W8 ^    super.buildActions();
' R! }, r2 H0 J! K   
    // Create the list of simulation actions. We put these in
    // an action group, because we want these actions to be
    // executed in a specific order, but these steps should
- ], v- u0 b1 O/ W9 |# T; a    // take no (simulated) time. The M(foo) means "The message
    // called <foo>". You can send a message To a particular
" C. W2 x5 E. O% U% Q# @  V    // object, or ForEach object in a collection.
+ s  f+ f: ?; Y" o+ u% g        
- G$ Z, e7 r+ Q! n. v    // Note we update the heatspace in two phases: first run
    // diffusion, then run "updateWorld" to actually enact the
    // changes the heatbugs have made. The ordering here is
' n+ @2 O( y6 X) D& M5 l    // significant!
5 k* P* B" U) p        
( z' k4 c, S1 h9 l) w/ v    // Note also, that with the additional
    // `randomizeHeatbugUpdateOrder' Boolean flag we can
    // randomize the order in which the bugs actually run
8 u$ V* o3 i) J4 }# D! b    // their step rule.  This has the effect of removing any
' H# M. g5 F: Z5 j! a/ T    // systematic bias in the iteration throught the heatbug
3 `0 x8 I! y0 e! g0 \    // list from timestep to timestep
        
    // By default, all `createActionForEach' modelActions have
" d8 y, J! z" j. N  q    // a default order of `Sequential', which means that the
) f/ V' x' n  G4 n1 Z6 r    // order of iteration through the `heatbugList' will be
! h) M/ Y# [% O- C: x6 ~% \    // identical (assuming the list order is not changed
    // indirectly by some other process).
( ?# |1 q, f! S% {" y& T( Q% l   
    modelActions = new ActionGroupImpl (getZone ());
. }7 W: p% T/ U2 H6 i6 U. c
# e* }' Y$ r  `    try {
# z5 A3 _$ C+ C3 G$ `$ P      modelActions.createActionTo$message
/ [# }' l0 u6 T9 X: A        (heat, new Selector (heat.getClass (), "stepRule", false));
    } catch (Exception e) {
      System.err.println ("Exception stepRule: " + e.getMessage ());
3 l0 V+ g* @; z! }: X% R    }

    try {
      Heatbug proto = (Heatbug) heatbugList.get (0);
9 [+ E4 p/ y; M8 s      Selector sel =
        new Selector (proto.getClass (), "heatbugStep", false);
2 q9 [# C4 d; c7 T( e1 l0 z) K      actionForEach =
8 V& k/ q/ y+ t, J, N. I        modelActions.createFActionForEachHomogeneous$call
        (heatbugList,
         new FCallImpl (this, proto, sel,
8 R; \/ k4 U, H( |* w                        new FArgumentsImpl (this, sel)));
    } catch (Exception e) {
* g4 N, \, O: U, b: [& r! p' L      e.printStackTrace (System.err);
    }
   
    syncUpdateOrder ();

    try {
0 r! W1 I8 o1 P: m: v& F/ m      modelActions.createActionTo$message
- j9 N3 y; d! U$ T5 r+ m+ I/ K6 r        (heat, new Selector (heat.getClass (), "updateLattice", false));
$ X  Z9 O) j0 X& q    } catch (Exception e) {
      System.err.println("Exception updateLattice: " + e.getMessage ());
    }
        
    // Then we create a schedule that executes the
    // modelActions. modelActions is an ActionGroup, by itself it
    // has no notion of time. In order to have it executed in
7 _9 ~! ~2 K% ]: Z* d7 G    // time, we create a Schedule that says to use the
% f1 i$ U8 l8 \/ U$ C    // modelActions ActionGroup at particular times.  This
3 O! I% c7 v4 _    // schedule has a repeat interval of 1, it will loop every
    // time step.  The action is executed at time 0 relative to
    // the beginning of the loop.

    // This is a simple schedule, with only one action that is
  L- V, @! t- B# V0 V    // just repeated every time. See jmousetrap for more
! b/ h+ q# ]# D    // complicated schedules.
9 `5 v) X8 h/ T. c1 ?$ \  
    modelSchedule = new ScheduleImpl (getZone (), 1);
    modelSchedule.at$createAction (0, modelActions);
( s, Y8 M& H# l: w        
  P9 |  ~. }$ y) z    return this;
  }