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：
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public Object buildActions () {
- e& }3 E0 C/ p1 E1 X    super.buildActions();
6 h3 q8 |) h% F3 Q. T4 n   
0 O' Q. }; t: S: p. ^    // 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
: {7 v! ?' ^  y' U3 E6 v7 O# Z4 [    // take no (simulated) time. The M(foo) means "The message
6 c5 U0 v8 j. D& X  v+ \    // called <foo>". You can send a message To a particular
1 m! e# a4 \, u' K- q  V5 y8 H    // object, or ForEach object in a collection.
        
    // Note we update the heatspace in two phases: first run
0 |9 d  ?  {( R3 L# I+ R, R% U    // diffusion, then run "updateWorld" to actually enact the
) ]; T: |, o* ^, t) Y1 X    // changes the heatbugs have made. The ordering here is
    // significant!
        
    // Note also, that with the additional
  R6 z' F3 ]* ?    // `randomizeHeatbugUpdateOrder' Boolean flag we can
7 ^# g; A. b+ s2 L3 j+ Q' H    // randomize the order in which the bugs actually run
! {1 t4 ]4 J& F+ M) h% X9 g    // their step rule.  This has the effect of removing any
    // systematic bias in the iteration throught the heatbug
/ V, b$ B1 d+ o! l' ~! R    // list from timestep to timestep
        
    // By default, all `createActionForEach' modelActions have
( I* v0 g. t# l. |  t    // a default order of `Sequential', which means that the
' m2 J# L! `+ j+ V: n  y1 O    // order of iteration through the `heatbugList' will be
0 J! x! ^9 K! i! e8 ?$ _# u( r    // identical (assuming the list order is not changed
    // indirectly by some other process).
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, g8 l9 F3 y4 h$ B' D) a# w    modelActions = new ActionGroupImpl (getZone ());

7 b% @6 S+ k) n$ h, V  ~$ H# J6 p    try {
      modelActions.createActionTo$message
        (heat, new Selector (heat.getClass (), "stepRule", false));
7 }1 D( F% v* R, `+ D    } catch (Exception e) {
      System.err.println ("Exception stepRule: " + e.getMessage ());
3 N2 c, ?# h/ y! }/ \    }

    try {
3 @1 f# E; z( b. p! w* n2 I# ^      Heatbug proto = (Heatbug) heatbugList.get (0);
, `2 k% ^. v) u+ _. G& r      Selector sel =
, _* [$ ~% z+ T! z, i2 X        new Selector (proto.getClass (), "heatbugStep", false);
: r* w3 Q: R& D      actionForEach =
. s6 o; K1 D( |& m3 o        modelActions.createFActionForEachHomogeneous$call
        (heatbugList,
/ z& k; e( R  d( g: V& e. m$ j9 N' x         new FCallImpl (this, proto, sel,
                        new FArgumentsImpl (this, sel)));
6 i% }8 T9 `( [, `# }    } catch (Exception e) {
" i+ X5 o! C0 P( q      e.printStackTrace (System.err);
    }
# S6 k, J* g( I& m  L8 }   
    syncUpdateOrder ();

    try {
+ A% d1 f2 w- A( o: c      modelActions.createActionTo$message
        (heat, new Selector (heat.getClass (), "updateLattice", false));
2 R+ o% [1 K9 k, O1 ?% K0 s, y. T. j. J    } catch (Exception e) {
      System.err.println("Exception updateLattice: " + e.getMessage ());
$ F8 K. h3 Q9 D) l  r+ [1 z/ ?    }
( Q/ b- V5 c0 t7 @% M! o* Z        
    // Then we create a schedule that executes the
    // modelActions. modelActions is an ActionGroup, by itself it
* T: ~& p& }, @    // has no notion of time. In order to have it executed in
    // time, we create a Schedule that says to use the
    // modelActions ActionGroup at particular times.  This
    // schedule has a repeat interval of 1, it will loop every
    // time step.  The action is executed at time 0 relative to
6 S: n5 c5 ^4 O0 ~) m+ z5 M    // the beginning of the loop.
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    // This is a simple schedule, with only one action that is
    // just repeated every time. See jmousetrap for more
% e1 a) ~& H4 z6 P9 Z    // complicated schedules.
  
    modelSchedule = new ScheduleImpl (getZone (), 1);
) r- @8 S5 U# r) L7 v  x    modelSchedule.at$createAction (0, modelActions);
        
    return this;
  }