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 M+ n$ e  O" L; L6 `
& l7 R0 V; X; R) h: m* T public Object buildActions () {
3 `' x& \  I* J+ P9 a( M    super.buildActions();
   
    // Create the list of simulation actions. We put these in
7 R  F1 S# F+ `# `' Q, W, {    // 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
8 {' q9 {. i' l  W! y    // called <foo>". You can send a message To a particular
* G, _! V4 {6 ~5 q" `) E    // object, or ForEach object in a collection.
/ W: V- U0 V% z, z        
    // 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
" A+ g3 G3 W# y    // significant!
: U3 I+ P4 Y" k        
    // Note also, that with the additional
) T# p4 }# s+ g" P3 q  M4 |0 e' z! U' ]9 W    // `randomizeHeatbugUpdateOrder' Boolean flag we can
    // randomize the order in which the bugs actually run
- o- ~+ U# \' Q- f. N8 {    // their step rule.  This has the effect of removing any
  E( \( s# c8 J$ I% r: ^! v    // systematic bias in the iteration throught the heatbug
+ [8 k% ~* v& C/ `- ?    // list from timestep to timestep
        
    // By default, all `createActionForEach' modelActions have
    // a default order of `Sequential', which means that the
    // order of iteration through the `heatbugList' will be
9 Q" ]- I. l0 ~3 y# f    // identical (assuming the list order is not changed
: `/ `  f: s8 e0 [8 G( ~" x    // indirectly by some other process).
   
    modelActions = new ActionGroupImpl (getZone ());
; p5 U4 S: X& G
    try {
* W+ p* J1 O  S# V( S2 Z      modelActions.createActionTo$message
        (heat, new Selector (heat.getClass (), "stepRule", false));
) e8 q# p& |9 @1 g2 B, H# V* ~) ?    } catch (Exception e) {
      System.err.println ("Exception stepRule: " + e.getMessage ());
% A' B5 d& m) ~% ?, E) a    }
6 K- ~7 m2 c! K  N% A' f
    try {
6 ?7 S, S6 v9 ~- w# X: E      Heatbug proto = (Heatbug) heatbugList.get (0);
      Selector sel =
        new Selector (proto.getClass (), "heatbugStep", false);
      actionForEach =
* H- ?, W- |7 P2 I- P  u        modelActions.createFActionForEachHomogeneous$call
        (heatbugList,
         new FCallImpl (this, proto, sel,
                        new FArgumentsImpl (this, sel)));
) I  E# v" B  u0 E8 f# }9 A    } catch (Exception e) {
      e.printStackTrace (System.err);
    }
6 x: D$ V. K) k5 x   
8 T+ Q( Z3 H7 G, w    syncUpdateOrder ();
3 f6 z! S! g+ |# v
    try {
      modelActions.createActionTo$message
  n2 m6 ?4 H1 Y        (heat, new Selector (heat.getClass (), "updateLattice", false));
    } catch (Exception e) {
0 R7 K) _$ @6 P: d      System.err.println("Exception updateLattice: " + e.getMessage ());
    }
        
    // Then we create a schedule that executes the
. V) w" t; Q& U3 c& ^! y" Q. s    // 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
    // modelActions ActionGroup at particular times.  This
    // schedule has a repeat interval of 1, it will loop every
- |/ V& w. [* G0 Z2 B/ R' [    // time step.  The action is executed at time 0 relative to
    // the beginning of the loop.
0 M: }$ z# v5 o" h
    // This is a simple schedule, with only one action that is
' ]) h% D- `& E  R    // just repeated every time. See jmousetrap for more
/ `+ B& W5 c- X9 v    // complicated schedules.
" d" h/ T, i: g  
    modelSchedule = new ScheduleImpl (getZone (), 1);
  ^- R+ m6 R% V" Q  S- D. j    modelSchedule.at$createAction (0, modelActions);
! t6 p4 Y* K9 g2 h9 R        
& b+ i  M* h7 ~7 T& D    return this;
  }