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 () {
    super.buildActions();
) E" z% @* j  I/ p- C' ^   
    // Create the list of simulation actions. We put these in
9 I  v" r* {* M' B    // an action group, because we want these actions to be
    // executed in a specific order, but these steps should
* E" R7 M! C5 ^! g    // take no (simulated) time. The M(foo) means "The message
5 `0 r  g% ?4 p" @( Q    // called <foo>". You can send a message To a particular
    // object, or ForEach object in a collection.
* v) f& f5 _9 H7 E0 W! H4 B        
    // Note we update the heatspace in two phases: first run
    // diffusion, then run "updateWorld" to actually enact the
2 U+ e0 a7 C1 ?3 W$ G8 e! T$ |    // changes the heatbugs have made. The ordering here is
    // significant!
7 F# \: V/ K$ B' ?4 i5 ~        
    // Note also, that with the additional
  L) m( x; ?5 B3 u! T2 F5 {% g* i: B0 l  t    // `randomizeHeatbugUpdateOrder' Boolean flag we can
1 y/ l: N) f: Q5 Q& r5 p    // randomize the order in which the bugs actually run
2 ~7 u6 L# c2 k. E' z1 Y    // their step rule.  This has the effect of removing any
    // systematic bias in the iteration throught the heatbug
    // list from timestep to timestep
! _% O; x8 J6 [% N1 n        
  X# d2 w$ Y6 b- U1 `5 L    // By default, all `createActionForEach' modelActions have
    // a default order of `Sequential', which means that the
2 J$ S8 G. c6 N  Q9 d5 l    // order of iteration through the `heatbugList' will be
4 V3 o- W& `& r; ]5 K0 ]" j+ B    // identical (assuming the list order is not changed
    // indirectly by some other process).
1 C1 J- O' |+ k' g7 H   
+ K1 H1 T0 C- `- z6 r$ D    modelActions = new ActionGroupImpl (getZone ());

    try {
      modelActions.createActionTo$message
2 U- B$ |1 K2 K# X: Z2 ^        (heat, new Selector (heat.getClass (), "stepRule", false));
; l9 X* M% K1 i1 K  `3 A( n    } catch (Exception e) {
* M" G; l3 T8 d, Q+ v" r# i      System.err.println ("Exception stepRule: " + e.getMessage ());
0 u# U2 y: }8 H) _    }

3 q0 f. Z, l- Y0 r! s    try {
6 T7 x2 P% F4 o8 ]      Heatbug proto = (Heatbug) heatbugList.get (0);
      Selector sel =
        new Selector (proto.getClass (), "heatbugStep", false);
      actionForEach =
+ C& s3 E: u+ K: M' z        modelActions.createFActionForEachHomogeneous$call
# i8 Q7 c$ M9 v        (heatbugList,
. ?& f' L- X% m( y         new FCallImpl (this, proto, sel,
                        new FArgumentsImpl (this, sel)));
" J) E9 a, O' |& K1 x0 E    } catch (Exception e) {
- u4 R! A' V: |0 R0 u# W      e.printStackTrace (System.err);
) `" G- x3 l5 L8 T    }
) |% A- ~1 d- t4 G, O/ ^  L9 F1 ~   
    syncUpdateOrder ();

0 @, A( F8 |7 l' R/ ^    try {
% [/ p4 o+ \" u  }- `  W, _      modelActions.createActionTo$message
        (heat, new Selector (heat.getClass (), "updateLattice", false));
, }  s0 F* @" U! T5 T) s1 @$ @4 l    } catch (Exception e) {
5 f5 ^/ \; g- e      System.err.println("Exception updateLattice: " + e.getMessage ());
3 U, N" o) D. N% e0 p# i) H    }
        
    // Then we create a schedule that executes the
    // modelActions. modelActions is an ActionGroup, by itself it
    // has no notion of time. In order to have it executed in
- `) x) ?+ g) L6 r. b; M    // time, we create a Schedule that says to use the
# u, T6 F) m2 d! A* u4 K    // modelActions ActionGroup at particular times.  This
. w9 a: F! w3 W    // schedule has a repeat interval of 1, it will loop every
    // time step.  The action is executed at time 0 relative to
9 q" }! w* {% i, t    // the beginning of the loop.

  f" n) e% X# l! S5 E5 l4 I# w    // This is a simple schedule, with only one action that is
7 q9 H& H8 p+ u9 c    // just repeated every time. See jmousetrap for more
6 ?! \1 @4 Z* H. E. e    // complicated schedules.
  
    modelSchedule = new ScheduleImpl (getZone (), 1);
    modelSchedule.at$createAction (0, modelActions);
        
    return this;
  }