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 () {
% j# n3 E6 b) R    super.buildActions();
0 K3 w2 e) H# X0 H- U7 @. D   
# h3 L- d: `7 V8 {1 q0 m    // Create the list of simulation actions. We put these in
, B2 \4 K$ c2 _0 _7 ~/ a8 }    // 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
    // object, or ForEach object in a collection.
, h, Z7 ~- n. Q: c5 g% }        
6 h6 d7 J5 y. b- D    // 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
2 p) [5 m; k/ b1 Y# z    // significant!
* _2 @7 x5 ^5 w+ G2 Y7 E7 A        
% L1 r' Q( S; c6 O2 L' n) k    // Note also, that with the additional
    // `randomizeHeatbugUpdateOrder' Boolean flag we can
    // randomize the order in which the bugs actually run
7 x% V9 T& O4 p6 f+ e& t: _    // their step rule.  This has the effect of removing any
    // systematic bias in the iteration throught the heatbug
    // list from timestep to timestep
, `% I& v) \- a) A% K$ c        
    // By default, all `createActionForEach' modelActions have
# g& Y/ \& }, ?* `0 d. t) C    // a default order of `Sequential', which means that the
  g  G  i- h6 S# b2 J$ l+ E    // order of iteration through the `heatbugList' will be
    // identical (assuming the list order is not changed
# U( c8 ?2 y+ w$ y1 g$ z5 L, E% t" t$ j4 }    // indirectly by some other process).
, l7 H$ y; j9 l" T   
    modelActions = new ActionGroupImpl (getZone ());
& C* i! [/ ^0 O7 A6 T2 y% S1 w
    try {
      modelActions.createActionTo$message
        (heat, new Selector (heat.getClass (), "stepRule", false));
+ d7 {/ N0 o5 a4 |    } catch (Exception e) {
      System.err.println ("Exception stepRule: " + e.getMessage ());
4 E- Z5 o  [' w% L9 J, C    }
! y) S; w- w+ J8 I0 U7 v* I
2 e  E) f* W0 ]& k) ?+ f    try {
      Heatbug proto = (Heatbug) heatbugList.get (0);
5 K0 s" Q3 o/ g5 M7 ]* k      Selector sel =
% X8 j& J8 B" Q        new Selector (proto.getClass (), "heatbugStep", false);
- u$ ?  S" O6 U/ K8 \) T8 V      actionForEach =
        modelActions.createFActionForEachHomogeneous$call
        (heatbugList,
& a# J1 P- `$ f         new FCallImpl (this, proto, sel,
                        new FArgumentsImpl (this, sel)));
1 V% w5 A/ u. [8 E4 s, u# Y    } catch (Exception e) {
      e.printStackTrace (System.err);
    }
   
    syncUpdateOrder ();

    try {
7 i6 Y( h( u; ~7 }- M      modelActions.createActionTo$message
% l, O, B5 G3 I2 h' M  g        (heat, new Selector (heat.getClass (), "updateLattice", false));
" r4 h, t4 I' k1 v2 B. V    } catch (Exception e) {
; i7 v8 K  b2 M( y0 y      System.err.println("Exception updateLattice: " + e.getMessage ());
    }
' s/ P8 [; ]6 F. `& `5 e: P) Y        
: O* O" l% R; _! P4 r    // Then we create a schedule that executes the
. l0 y$ G% a- |8 C# P/ ^0 I" `    // modelActions. modelActions is an ActionGroup, by itself it
    // has no notion of time. In order to have it executed in
+ @' R) l7 i  x) j    // 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
* h3 j! R5 u! c    // time step.  The action is executed at time 0 relative to
    // the beginning of the loop.

    // This is a simple schedule, with only one action that is
9 ?  f. d; {. W" C    // just repeated every time. See jmousetrap for more
    // complicated schedules.
  
    modelSchedule = new ScheduleImpl (getZone (), 1);
. K! B) A/ i3 a- |* v4 g    modelSchedule.at$createAction (0, modelActions);
8 N1 }! \2 I; @2 i% J) e' V        
    return this;
  [4 T7 ?* M1 I2 h, |( I. ]  X  }