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 () {
$ s& H, ]7 N5 E9 ~1 \0 K3 h9 j    super.buildActions();
* `. p" D6 h, ]2 b, v   
    // 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
    // called <foo>". You can send a message To a particular
    // object, or ForEach object in a collection.
, k; c3 ~. q4 w1 k1 k' }. X% u# x2 |& _% }        
    // Note we update the heatspace in two phases: first run
    // diffusion, then run "updateWorld" to actually enact the
) k0 X5 _$ f9 @: v. `    // changes the heatbugs have made. The ordering here is
' E$ L; K" v  \# t. m. g    // significant!
8 F. x% A6 v( N        
& R0 ?" s+ ~& K' {% m/ i% v    // Note also, that with the additional
    // `randomizeHeatbugUpdateOrder' Boolean flag we can
( U$ {. P0 x& J5 j& ~8 o$ d    // randomize the order in which the bugs actually run
( O; x$ [( U' @+ n# Y' n    // their step rule.  This has the effect of removing any
& V* ?& n4 t2 }9 V+ V! L% h. ?! u    // systematic bias in the iteration throught the heatbug
: z+ o" i2 ~0 `% Y- A    // list from timestep to timestep
8 T: |: E6 _9 r        
    // By default, all `createActionForEach' modelActions have
# y8 w$ w6 q# r1 H! L. n- @% ?    // 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
    // indirectly by some other process).
/ X: B' C4 l/ p: L, M2 M. H9 C   
    modelActions = new ActionGroupImpl (getZone ());
: I+ T6 c* g* ^, V) S/ \2 p& `
) P7 ]( }, `5 y+ e5 B. d3 m* d1 J8 D    try {
. E% `2 ~6 Y6 _# q      modelActions.createActionTo$message
        (heat, new Selector (heat.getClass (), "stepRule", false));
% C$ u8 Y+ z- ^. W    } catch (Exception e) {
      System.err.println ("Exception stepRule: " + e.getMessage ());
    }
  C6 ], p# M% k% e2 X4 Z
( m$ f  E* P; z2 ~8 b. o    try {
      Heatbug proto = (Heatbug) heatbugList.get (0);
  f: Y# K5 k' w$ n; d      Selector sel =
2 O6 \+ a% K5 J8 ]' ?% p7 d        new Selector (proto.getClass (), "heatbugStep", false);
      actionForEach =
        modelActions.createFActionForEachHomogeneous$call
) m8 ~5 J+ e' s3 X' V: c# x) u& O3 @        (heatbugList,
         new FCallImpl (this, proto, sel,
                        new FArgumentsImpl (this, sel)));
    } catch (Exception e) {
      e.printStackTrace (System.err);
    }
   
+ ~0 ?& n8 J6 j; G: |    syncUpdateOrder ();

    try {
      modelActions.createActionTo$message
        (heat, new Selector (heat.getClass (), "updateLattice", false));
    } catch (Exception e) {
      System.err.println("Exception updateLattice: " + e.getMessage ());
    }
. w1 F9 y/ k0 Y        
- y% r/ w7 `  s$ Y" j    // Then we create a schedule that executes the
    // modelActions. modelActions is an ActionGroup, by itself it
0 L- Q6 e$ a* ?  N% T    // has no notion of time. In order to have it executed in
0 X( s8 i1 z- ~  t' f    // 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
! V9 O+ \( S8 z! J2 E    // time step.  The action is executed at time 0 relative to
) z0 f3 F. Q7 u9 n6 o! q) m; q: @7 b    // the beginning of the loop.

    // 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);
  H9 a- c8 O$ R# Z( g: N& y: L8 {    modelSchedule.at$createAction (0, modelActions);
        
- r: D5 }5 {' V& P+ w$ @    return this;
3 }  [/ S3 Z% Q, e$ [( f- H4 m  }