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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' X$ B2 v0 T; x public Object buildActions () {
    super.buildActions();
! N1 }1 d4 V3 F' y3 }; }  r, z   
! \# D6 q1 ~; z0 L    // Create the list of simulation actions. We put these in
, v; J  a: S1 ]) [    // an action group, because we want these actions to be
    // executed in a specific order, but these steps should
3 ?6 o$ }2 `  W+ t3 U. }8 x& ^    // 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.
% h; Z1 N5 e8 O1 T% c' {7 G        
    // Note we update the heatspace in two phases: first run
$ A$ T$ I* @/ A- }1 u+ S1 t    // diffusion, then run "updateWorld" to actually enact the
+ }' E& W; }3 U1 Z' S/ \& g% z5 g, a    // changes the heatbugs have made. The ordering here is
    // significant!
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: d: `5 |4 v8 G" n    // 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
. }8 Q" }2 k9 v6 S6 v7 z! l    // list from timestep to timestep
- k* E# {4 m1 }        
( s6 V; t9 }5 l4 y2 M# }+ f    // By default, all `createActionForEach' modelActions have
    // 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
# B1 `: @1 [' c; [' e    // indirectly by some other process).
   
- b5 x, @+ G& P; V- P    modelActions = new ActionGroupImpl (getZone ());

+ G+ J. Z2 _% k    try {
0 G' Q% @. N; e1 A- @      modelActions.createActionTo$message
        (heat, new Selector (heat.getClass (), "stepRule", false));
# d& v# v2 r7 {" k# @0 D    } catch (Exception e) {
      System.err.println ("Exception stepRule: " + e.getMessage ());
6 L" O( S) |) h7 N1 V    }
+ I3 A# u; d% F0 [7 Q; h, n
- {6 p9 n3 G7 j0 e# l    try {
$ F& _7 C% O  F8 {9 [      Heatbug proto = (Heatbug) heatbugList.get (0);
      Selector sel =
        new Selector (proto.getClass (), "heatbugStep", false);
      actionForEach =
        modelActions.createFActionForEachHomogeneous$call
        (heatbugList,
         new FCallImpl (this, proto, sel,
                        new FArgumentsImpl (this, sel)));
    } catch (Exception e) {
      e.printStackTrace (System.err);
# u" ^* a) g2 x/ d; e' ~% G    }
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) f( u% o8 [' X5 L6 g7 J- f5 n    syncUpdateOrder ();
% s& D' w/ i2 h7 i+ a
    try {
      modelActions.createActionTo$message
$ C# V& x6 c/ I6 J- C  N8 F        (heat, new Selector (heat.getClass (), "updateLattice", false));
8 n. A/ C) ?, u# W; R    } catch (Exception e) {
      System.err.println("Exception updateLattice: " + e.getMessage ());
$ H5 m" t9 i  T- g: N/ s% o    }
        
- r  w& \, [6 S  A/ `    // Then we create a schedule that executes the
+ D- m/ H! j. W    // modelActions. modelActions is an ActionGroup, by itself it
$ Z3 \9 B4 v) l( Q8 m/ g. x4 K    // 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
1 F" }" O0 Q) D- t$ Z    // time step.  The action is executed at time 0 relative to
8 J+ i4 Z2 ]% l9 X. Q5 E, L, \; {    // the beginning of the loop.

. c/ H% M) o9 d    // This is a simple schedule, with only one action that is
, F. {) U+ i. b8 G5 d    // just repeated every time. See jmousetrap for more
' }  e) c0 Q9 A    // complicated schedules.
  
+ @8 H6 {5 _7 z, D5 G& e    modelSchedule = new ScheduleImpl (getZone (), 1);
  l+ s2 k! N3 B- [    modelSchedule.at$createAction (0, modelActions);
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    return this;
( N# Z2 w  \5 M7 T! r, Q# |  }