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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8 p# B, p; P" y1 _% d public Object buildActions () {
    super.buildActions();
   
+ g4 y: W% ^: d0 V; K. }* A; H    // Create the list of simulation actions. We put these in
    // an action group, because we want these actions to be
& R  e  d! t9 Z6 G+ Q0 H: h    // executed in a specific order, but these steps should
; K. d# i/ _9 h5 V6 B- }    // take no (simulated) time. The M(foo) means "The message
% U& O" N3 ]3 P: r( u: Z( r    // called <foo>". You can send a message To a particular
    // object, or ForEach object in a collection.
        
    // Note we update the heatspace in two phases: first run
" b8 V0 Y- k+ M2 T    // diffusion, then run "updateWorld" to actually enact the
    // changes the heatbugs have made. The ordering here is
    // significant!
, j5 A, \. O( x1 v) f        
2 T6 j- m$ y4 b- m( `    // Note also, that with the additional
    // `randomizeHeatbugUpdateOrder' Boolean flag we can
) B5 x: l9 t2 u0 J% z8 U) h    // randomize the order in which the bugs actually run
3 b7 n) n1 ]& T) l$ Y3 t9 M    // their step rule.  This has the effect of removing any
    // systematic bias in the iteration throught the heatbug
" i9 e2 a/ Z0 [3 R# P7 ~    // list from timestep to timestep
: m2 Q: {' E7 F( z  c3 h        
    // By default, all `createActionForEach' modelActions have
    // a default order of `Sequential', which means that the
/ B7 Z- E$ U% X8 |2 C9 B    // order of iteration through the `heatbugList' will be
" n% j/ i7 a6 Y9 w    // identical (assuming the list order is not changed
    // indirectly by some other process).
   
    modelActions = new ActionGroupImpl (getZone ());
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    try {
1 G: y: L- q  i. O: D      modelActions.createActionTo$message
        (heat, new Selector (heat.getClass (), "stepRule", false));
( F! Y+ i4 t  Q& b    } catch (Exception e) {
- V$ y2 n" q  K3 Z      System.err.println ("Exception stepRule: " + e.getMessage ());
    }
  w3 L) \' x4 W5 u% W% \' q
2 z3 A* u3 j/ [0 E1 `    try {
2 _5 u( Q( P) _8 A, s' u' i      Heatbug proto = (Heatbug) heatbugList.get (0);
      Selector sel =
        new Selector (proto.getClass (), "heatbugStep", false);
5 Y( X2 d# ]3 E$ @  H4 U- A0 m+ p      actionForEach =
9 u1 B% }0 G; m) L9 Y& ]$ Y. f3 v        modelActions.createFActionForEachHomogeneous$call
        (heatbugList,
4 T0 d+ b$ x& a1 b         new FCallImpl (this, proto, sel,
                        new FArgumentsImpl (this, sel)));
    } catch (Exception e) {
% e3 Y4 ^. o% S2 T, c$ W4 ~      e.printStackTrace (System.err);
3 k" s6 p: l, t$ ?    }
   
    syncUpdateOrder ();

    try {
4 C) {2 G1 Q. `9 Q9 _      modelActions.createActionTo$message
        (heat, new Selector (heat.getClass (), "updateLattice", false));
% |/ p9 b+ y6 o' M# b    } catch (Exception e) {
8 {1 J: |7 Q: c$ v! E7 \' x6 \      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
$ U) J, d: A  f8 }( A6 k    // time, we create a Schedule that says to use the
3 D, H0 i8 o; l* F    // modelActions ActionGroup at particular times.  This
7 J: G; {; P6 l7 f- ~9 J    // schedule has a repeat interval of 1, it will loop every
$ C" N) Q7 u. d* e, a    // time step.  The action is executed at time 0 relative to
/ G0 Q& a+ w* N& L    // the beginning of the loop.

' D4 D" V, W5 Z    // This is a simple schedule, with only one action that is
6 x9 M/ A3 w" z    // just repeated every time. See jmousetrap for more
* A* w2 r# p$ d    // complicated schedules.
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    modelSchedule = new ScheduleImpl (getZone (), 1);
    modelSchedule.at$createAction (0, modelActions);
. ^' t+ c+ `0 ?4 {# H& j        
    return this;
2 J& `, Y* Y5 d8 v5 v. ^6 T  }