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 () {
    super.buildActions();
. r$ H6 W. o/ a9 J# h   
    // Create the list of simulation actions. We put these in
    // an action group, because we want these actions to be
( h: v, v6 u6 a( p3 q    // executed in a specific order, but these steps should
! G/ O+ {0 a5 Y: X8 \, F  G0 K    // take no (simulated) time. The M(foo) means "The message
    // called <foo>". You can send a message To a particular
7 g* ]2 K. G7 N6 B4 R8 M$ ]: p    // object, or ForEach object in a collection.
/ L- Y/ q3 d& |+ v        
/ G$ @9 F1 i# ~    // Note we update the heatspace in two phases: first run
    // diffusion, then run "updateWorld" to actually enact the
: }  c8 I! H! R6 r& C    // changes the heatbugs have made. The ordering here is
    // significant!
: S% Z+ z7 O2 I; y2 d; K# _        
7 ]6 o$ k; c; X+ N    // Note also, that with the additional
1 C4 w  }. ]' P    // `randomizeHeatbugUpdateOrder' Boolean flag we can
    // randomize the order in which the bugs actually run
    // their step rule.  This has the effect of removing any
% O$ l* a$ S2 v( h4 s; M    // systematic bias in the iteration throught the heatbug
+ `! J# j6 o% ?7 N# ?    // list from timestep to timestep
        
    // By default, all `createActionForEach' modelActions have
    // a default order of `Sequential', which means that the
' Z3 z' l9 I$ c" {! c    // order of iteration through the `heatbugList' will be
    // identical (assuming the list order is not changed
, y1 x8 k6 X/ B& O# a    // indirectly by some other process).
  Q! s* f4 `$ t* H* A   
% f4 S7 ?/ N2 D& m    modelActions = new ActionGroupImpl (getZone ());

' S' F' W8 F5 K0 u! }  c! |    try {
# \' d) H3 Y  h4 a# R      modelActions.createActionTo$message
        (heat, new Selector (heat.getClass (), "stepRule", false));
    } catch (Exception e) {
# q/ ^- A% x; y5 J- c) b+ T      System.err.println ("Exception stepRule: " + e.getMessage ());
    }
; u) ?9 ]6 E1 t8 w0 m7 }
    try {
      Heatbug proto = (Heatbug) heatbugList.get (0);
$ ~3 a/ Q2 h% j      Selector sel =
        new Selector (proto.getClass (), "heatbugStep", false);
      actionForEach =
        modelActions.createFActionForEachHomogeneous$call
        (heatbugList,
         new FCallImpl (this, proto, sel,
                        new FArgumentsImpl (this, sel)));
+ V7 |# M! _0 K+ _6 q3 _    } catch (Exception e) {
      e.printStackTrace (System.err);
    }
   
    syncUpdateOrder ();
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    try {
: U6 \% `# a) |6 D4 F* w( t0 m0 o      modelActions.createActionTo$message
        (heat, new Selector (heat.getClass (), "updateLattice", false));
    } catch (Exception e) {
      System.err.println("Exception updateLattice: " + e.getMessage ());
# a  i4 K* C+ H3 p0 U1 ~8 j    }
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    // Then we create a schedule that executes the
    // modelActions. modelActions is an ActionGroup, by itself it
    // has no notion of time. In order to have it executed in
) ?% e3 `7 Q5 D2 g- B  e" a3 s( \+ v    // time, we create a Schedule that says to use the
4 [/ d) ]1 B& n; N9 s& H    // modelActions ActionGroup at particular times.  This
1 j/ s  N; e$ R" l& W5 m. C2 v    // schedule has a repeat interval of 1, it will loop every
) ^& W* P- {2 f$ u0 f    // time step.  The action is executed at time 0 relative to
! ?% l) e. X* n5 z9 ^) o( N; |    // 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
    // complicated schedules.
  
    modelSchedule = new ScheduleImpl (getZone (), 1);
    modelSchedule.at$createAction (0, modelActions);
$ |" t5 X- ~/ \. x$ C        
* s$ R# B- D# O' r    return this;
7 \6 n7 T7 W  ~1 K; V  }