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：
3 j# X1 l2 r& f. u( H9 _
public Object buildActions () {
    super.buildActions();
   
; ]$ _1 X( H) p1 ^+ B0 N7 i- _& g    // Create the list of simulation actions. We put these in
    // an action group, because we want these actions to be
' }2 G, @* Q, H0 \$ n5 d    // 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.
2 W2 r+ U8 C3 L( @& t        
    // Note we update the heatspace in two phases: first run
6 Q: L* `$ }5 _- q; l4 U    // diffusion, then run "updateWorld" to actually enact the
. [3 D! n& H9 `0 J9 L5 g    // changes the heatbugs have made. The ordering here is
4 ^9 W; ?, [) n6 b0 N! n  i4 f    // significant!
' ~, C) |: N. b! L! V& w+ G        
    // Note also, that with the additional
) V- n1 |8 F6 e' |; D. l$ Y0 p    // `randomizeHeatbugUpdateOrder' Boolean flag we can
    // randomize the order in which the bugs actually run
: r4 O8 Q* b3 z& Q# n. W; A    // their step rule.  This has the effect of removing any
! q6 F0 K0 F/ D( o$ G2 V! x, R    // systematic bias in the iteration throught the heatbug
    // list from timestep to timestep
        
+ y9 _0 p% |8 Z; `/ r    // 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
    // indirectly by some other process).
& J6 q2 c  A+ x- V" B8 \  {   
( {1 T7 k1 |: Y/ o    modelActions = new ActionGroupImpl (getZone ());

& o0 o6 E1 i# C8 t8 D+ d    try {
8 d, k; @% P/ ^! O      modelActions.createActionTo$message
! A' t% c( M, N; l* R) l# q        (heat, new Selector (heat.getClass (), "stepRule", false));
* t4 C5 ]4 F: P% J# m9 w  t. N2 ]    } catch (Exception e) {
      System.err.println ("Exception stepRule: " + e.getMessage ());
    }

    try {
      Heatbug proto = (Heatbug) heatbugList.get (0);
. q- O' T* h( _# g      Selector sel =
        new Selector (proto.getClass (), "heatbugStep", false);
      actionForEach =
/ h/ k9 d! b; G5 K. W. A5 N        modelActions.createFActionForEachHomogeneous$call
        (heatbugList,
         new FCallImpl (this, proto, sel,
+ W5 |: _* C8 `* I                        new FArgumentsImpl (this, sel)));
    } catch (Exception e) {
      e.printStackTrace (System.err);
    }
   
, a4 B9 |3 e9 A2 D  o- [+ O$ s    syncUpdateOrder ();

    try {
4 ^! `( M, D3 z. C: c# W      modelActions.createActionTo$message
        (heat, new Selector (heat.getClass (), "updateLattice", false));
    } catch (Exception e) {
6 a5 u- P+ P0 _& T. a3 V& @4 X' l      System.err.println("Exception updateLattice: " + e.getMessage ());
    }
        
    // Then we create a schedule that executes the
+ ^9 k! {6 ~$ F7 B: }8 {) m6 I6 `    // 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
. @8 j7 \. I8 Y7 Y; J8 ^4 y+ [    // modelActions ActionGroup at particular times.  This
; o0 [2 w" W) g& R4 l7 M    // schedule has a repeat interval of 1, it will loop every
4 b: G4 g% o5 L  q6 l  T+ p    // time step.  The action is executed at time 0 relative to
    // the beginning of the loop.
  \0 p& e8 `! G! ]8 p, }* r3 l
    // This is a simple schedule, with only one action that is
    // just repeated every time. See jmousetrap for more
4 d/ m& Q4 l4 v. t7 |) A8 E    // complicated schedules.
  
$ _+ B5 {6 a" F5 ~3 ^, ?+ c    modelSchedule = new ScheduleImpl (getZone (), 1);
    modelSchedule.at$createAction (0, modelActions);
        
( ]3 L+ P, g. n0 Y% K6 g+ i    return this;
  }