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：

9 K+ a5 b, L8 Z" W+ L4 K public Object buildActions () {
( U  G; i2 [& k) c# C$ s    super.buildActions();
$ x# d1 z) B$ R8 m7 D   
    // 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
    // take no (simulated) time. The M(foo) means "The message
' }) s  C8 M8 Z* m0 G% I) L8 u    // called <foo>". You can send a message To a particular
5 P) H2 `% |* C: Q: [) B    // object, or ForEach object in a collection.
4 ?: U8 s0 e- b! ?# U5 e8 w' g        
    // Note we update the heatspace in two phases: first run
* {% `5 X1 W4 t8 Z- Q    // diffusion, then run "updateWorld" to actually enact the
    // changes the heatbugs have made. The ordering here is
    // significant!
        
$ [; ]+ E3 y/ W    // Note also, that with the additional
    // `randomizeHeatbugUpdateOrder' Boolean flag we can
    // randomize the order in which the bugs actually run
( e$ `" h( [$ Q# T) D    // their step rule.  This has the effect of removing any
- ]& W, x) N& F+ r    // systematic bias in the iteration throught the heatbug
- Q( O5 K9 N2 a& ^( t: R* \    // list from timestep to timestep
- N% T) p/ j6 n! O" v        
9 P* w! B( d8 o7 y% u, j5 k    // By default, all `createActionForEach' modelActions have
    // a default order of `Sequential', which means that the
    // order of iteration through the `heatbugList' will be
    // identical (assuming the list order is not changed
) t- u3 p6 r6 k' c% b9 G7 m5 V$ R  h    // indirectly by some other process).
   
    modelActions = new ActionGroupImpl (getZone ());

0 @: T( ^0 m$ M6 j    try {
) K4 _  {0 I' h4 ~, g      modelActions.createActionTo$message
5 }/ k: l4 Y) j' o" _3 v5 E        (heat, new Selector (heat.getClass (), "stepRule", false));
    } catch (Exception e) {
2 Y. j; j7 {: d3 ^      System.err.println ("Exception stepRule: " + e.getMessage ());
    }

    try {
      Heatbug proto = (Heatbug) heatbugList.get (0);
; }% z, l$ ]# {1 B+ t      Selector sel =
        new Selector (proto.getClass (), "heatbugStep", false);
) C. l) D9 T0 l0 L* G$ g$ b      actionForEach =
, H, u  X$ W4 G7 T; S; P- y+ F        modelActions.createFActionForEachHomogeneous$call
        (heatbugList,
         new FCallImpl (this, proto, sel,
9 o: {2 O8 t; G) T; X# m0 R0 X; J                        new FArgumentsImpl (this, sel)));
    } catch (Exception e) {
      e.printStackTrace (System.err);
    }
   
    syncUpdateOrder ();
; Y) P8 H+ `9 p1 p# n
8 I" X. i9 J2 W    try {
      modelActions.createActionTo$message
" W! n9 y1 |. E' n& c0 A' b8 g" o        (heat, new Selector (heat.getClass (), "updateLattice", false));
9 j9 r" `2 u0 b) I    } catch (Exception e) {
, K1 G& Y* W; t: s      System.err.println("Exception updateLattice: " + e.getMessage ());
    }
) ~, E: u/ ?% V3 V$ N        
9 ~  @+ z( y6 N; x& K7 w    // 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
    // time, we create a Schedule that says to use the
( A; z9 }" Y9 z  o6 a2 C) s1 h    // modelActions ActionGroup at particular times.  This
8 m7 V& t- C/ V1 v% k) e' n    // schedule has a repeat interval of 1, it will loop every
    // time step.  The action is executed at time 0 relative to
3 j( p' o7 |- }5 P4 z    // the beginning of the loop.
; B% X4 \+ N" f" g; B
    // This is a simple schedule, with only one action that is
    // just repeated every time. See jmousetrap for more
  X' e2 @& `: ^  t) ~0 |    // complicated schedules.
  
    modelSchedule = new ScheduleImpl (getZone (), 1);
( h: D1 V0 L: R4 M) B    modelSchedule.at$createAction (0, modelActions);
4 @6 j" h1 v% H9 ?7 P& ^) n& [# P        
    return this;
  }