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：
0 K, t' R! T8 b0 p0 k+ @4 E
public Object buildActions () {
+ s8 S( q* B, O( a/ A# j    super.buildActions();
   
4 O! Q: x4 @$ K    // Create the list of simulation actions. We put these in
( K+ L5 ]$ G/ l: L5 |    // 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
! G1 O+ g6 s7 G! Q2 r& T3 n) ~    // object, or ForEach object in a collection.
        
    // Note we update the heatspace in two phases: first run
    // diffusion, then run "updateWorld" to actually enact the
    // changes the heatbugs have made. The ordering here is
& l. g! r% s2 W8 }6 ]4 r$ P    // significant!
6 @: \5 k/ X# X1 w5 e2 q9 K2 _        
4 s' H4 G0 R( {& P) e2 x    // 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
    // systematic bias in the iteration throught the heatbug
    // list from timestep to timestep
( ~& J& ^5 I3 [) y5 v2 C% _. Q        
9 e9 F. X  o+ R. g* @    // By default, all `createActionForEach' modelActions have
6 l' A. H4 l& {+ K- B1 O6 q    // a default order of `Sequential', which means that the
5 M3 M, z4 |6 |8 L1 f- o3 k    // order of iteration through the `heatbugList' will be
    // identical (assuming the list order is not changed
    // indirectly by some other process).
   
% J) }) ?: R" X" S% M% l4 P    modelActions = new ActionGroupImpl (getZone ());

/ t7 I* I- C, B; F0 f    try {
! E3 O6 X6 j3 e2 ?; y$ D      modelActions.createActionTo$message
$ N$ e: U+ e1 h) _        (heat, new Selector (heat.getClass (), "stepRule", false));
4 l' P5 _4 k! U5 x5 p' b    } catch (Exception e) {
2 g. R1 n8 ?9 t- h      System.err.println ("Exception stepRule: " + e.getMessage ());
7 G% [: l- d  |9 i+ E3 J! k$ L    }

: b2 K4 A$ P- c    try {
      Heatbug proto = (Heatbug) heatbugList.get (0);
& N1 ]; n  S% I8 }6 l' h      Selector sel =
        new Selector (proto.getClass (), "heatbugStep", false);
      actionForEach =
        modelActions.createFActionForEachHomogeneous$call
  X, G% ^" j! ^0 A* L        (heatbugList,
         new FCallImpl (this, proto, sel,
                        new FArgumentsImpl (this, sel)));
" E( H$ `4 O" t3 y: ]$ e1 r* |    } catch (Exception e) {
      e.printStackTrace (System.err);
    }
* d+ D  l) E7 K7 K, E2 T8 E) L' `   
5 c# O. q1 S$ M. g    syncUpdateOrder ();
1 f8 c4 e5 I, S7 a
! D) X8 R3 n) P; W. y; J6 F5 Y" j    try {
      modelActions.createActionTo$message
+ z  g) x% |# ]2 b        (heat, new Selector (heat.getClass (), "updateLattice", false));
    } catch (Exception e) {
% e% B0 K0 L) P      System.err.println("Exception updateLattice: " + e.getMessage ());
    }
* F1 g8 c& E& Q        
    // Then we create a schedule that executes the
, [7 R$ F. G% W    // modelActions. modelActions is an ActionGroup, by itself it
    // has no notion of time. In order to have it executed in
9 J8 K4 ^& @, S* f7 M) E1 g    // time, we create a Schedule that says to use the
) a" B% r$ Y8 Y; C" j+ t7 ]    // modelActions ActionGroup at particular times.  This
    // schedule has a repeat interval of 1, it will loop every
9 [* i& g1 p! u. ~" D* q    // time step.  The action is executed at time 0 relative to
    // the beginning of the loop.
( M; Q' L1 i& l; T( c
    // This is a simple schedule, with only one action that is
- u9 x  ~  h. m5 U' [    // just repeated every time. See jmousetrap for more
    // complicated schedules.
/ e2 ?1 C' _& v2 x* ^. T  
    modelSchedule = new ScheduleImpl (getZone (), 1);
    modelSchedule.at$createAction (0, modelActions);
        
6 T" r4 b. }( K4 K5 ]. \    return this;
  }