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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, m& U- _! y3 W3 X2 r: g0 `* e public Object buildActions () {
0 q, ]$ y7 p8 X# F9 d# D    super.buildActions();
   
1 _/ y6 e% F0 z: V- Z6 g    // Create the list of simulation actions. We put these in
    // an action group, because we want these actions to be
9 x: p, u1 E. v: u+ q6 Q0 K) n    // executed in a specific order, but these steps should
$ T/ l6 F( L( y! p    // take no (simulated) time. The M(foo) means "The message
* L' A# o3 M! D4 H, Z7 I    // called <foo>". You can send a message To a particular
    // object, or ForEach object in a collection.
. K! w4 b! q$ D2 a        
$ Z" {& c/ V! }: J    // Note we update the heatspace in two phases: first run
6 K1 l; `# |7 O3 m    // diffusion, then run "updateWorld" to actually enact the
    // changes the heatbugs have made. The ordering here is
    // significant!
        
    // Note also, that with the additional
    // `randomizeHeatbugUpdateOrder' Boolean flag we can
    // randomize the order in which the bugs actually run
    // their step rule.  This has the effect of removing any
7 W( F" G0 ~& c5 P; P    // systematic bias in the iteration throught the heatbug
2 h: z) S5 S- e0 `8 j( R$ ?    // list from timestep to timestep
        
" y7 n9 g/ O6 ^; |- Z/ m+ i    // By default, all `createActionForEach' modelActions have
    // a default order of `Sequential', which means that the
4 i3 Q, U0 |2 \; j# d% a: M    // order of iteration through the `heatbugList' will be
    // identical (assuming the list order is not changed
1 ]2 [* V. `' ^* S    // indirectly by some other process).
" l' H* L8 C* D* P3 o   
    modelActions = new ActionGroupImpl (getZone ());

# i* r" ?; E0 }    try {
      modelActions.createActionTo$message
* L  A8 S. y5 `1 p) ]        (heat, new Selector (heat.getClass (), "stepRule", false));
( T$ [: `, ?4 m' P7 m    } catch (Exception e) {
      System.err.println ("Exception stepRule: " + e.getMessage ());
2 R$ G& y4 _' d% t5 J* q    }
6 x1 [$ D2 G9 H2 B: \! g; Z1 S
    try {
      Heatbug proto = (Heatbug) heatbugList.get (0);
      Selector sel =
/ A0 t) w; h, j3 `0 g! q7 ?; \) }        new Selector (proto.getClass (), "heatbugStep", false);
  d2 x7 i% p! |. M7 I      actionForEach =
( i* }2 r& n: a) w        modelActions.createFActionForEachHomogeneous$call
        (heatbugList,
6 ]$ A, L3 O' B/ u  T3 U         new FCallImpl (this, proto, sel,
  x& P0 L% }+ I5 a                        new FArgumentsImpl (this, sel)));
7 N0 \4 _5 T6 _- G+ i    } catch (Exception e) {
      e.printStackTrace (System.err);
    }
: s8 w% w" i0 `! h   
' y  ^2 h& C1 {) s: P    syncUpdateOrder ();
3 t9 w3 V' o& v
0 a- z% E8 @# V7 \0 H  k+ e    try {
# [# W" M$ C9 A' q& }- E      modelActions.createActionTo$message
: F. ?) V, E5 m: n% z2 h( g( m        (heat, new Selector (heat.getClass (), "updateLattice", false));
    } catch (Exception e) {
; n4 n+ U! c, n      System.err.println("Exception updateLattice: " + e.getMessage ());
& ^" @0 V. ^  ^    }
+ ~! k& q. Q# h  {3 i        
$ D/ g5 z! o/ m    // Then we create a schedule that executes the
    // modelActions. modelActions is an ActionGroup, by itself it
; @) f# i+ r% b; E6 [  |    // 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
    // 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.
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    modelSchedule = new ScheduleImpl (getZone (), 1);
0 d$ i# K# c* C% O/ r" ]- o/ e. ?    modelSchedule.at$createAction (0, modelActions);
3 i5 ~1 q/ g3 n4 P; p6 j        
    return this;
  }