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：
. s) H3 y8 U5 @) K# b& V
public Object buildActions () {
, F8 O9 n1 X9 Z& Q  K    super.buildActions();
8 @/ b7 k8 ]  }' K$ f/ J! l   
7 o, ^5 t1 x9 `4 f- V/ ?4 z+ o    // Create the list of simulation actions. We put these in
    // an action group, because we want these actions to be
& P( {+ ~* S9 Z5 A' w8 l5 C    // executed in a specific order, but these steps should
9 u# f- @( D; w& M    // take no (simulated) time. The M(foo) means "The message
% T* y9 l9 \! g; A) S" }    // called <foo>". You can send a message To a particular
5 P5 B' w- s2 |. n3 ]    // object, or ForEach object in a collection.
        
    // 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
& p' S1 }0 f8 N    // significant!
3 y4 j  S4 \  `6 N* K0 M5 {9 _        
$ Y/ a- N9 b* }9 _    // Note also, that with the additional
    // `randomizeHeatbugUpdateOrder' Boolean flag we can
    // randomize the order in which the bugs actually run
. B5 ~$ b. j' ~* n' h; a+ l    // their step rule.  This has the effect of removing any
    // systematic bias in the iteration throught the heatbug
. r6 Y. v5 O3 \    // list from timestep to timestep
        
2 v3 `4 R  M) e  B    // By default, all `createActionForEach' modelActions have
    // a default order of `Sequential', which means that the
! J8 S+ E8 [6 z" |. Q    // order of iteration through the `heatbugList' will be
    // identical (assuming the list order is not changed
    // indirectly by some other process).
   
7 m, P; t* n1 B3 p5 c& F    modelActions = new ActionGroupImpl (getZone ());

    try {
      modelActions.createActionTo$message
$ h6 n$ j; t0 ]3 k        (heat, new Selector (heat.getClass (), "stepRule", false));
    } catch (Exception e) {
      System.err.println ("Exception stepRule: " + e.getMessage ());
    }

- K- I/ J0 L6 @+ W( s; m) @9 Y    try {
* H+ L, M* t# W- i      Heatbug proto = (Heatbug) heatbugList.get (0);
/ m7 q: q4 h% U      Selector sel =
        new Selector (proto.getClass (), "heatbugStep", false);
* {! C$ q) g1 ?0 [" Z      actionForEach =
! b) Y/ B$ `& C4 g' g        modelActions.createFActionForEachHomogeneous$call
        (heatbugList,
3 U+ j% J  U5 B! c         new FCallImpl (this, proto, sel,
                        new FArgumentsImpl (this, sel)));
: ]0 [0 z' _0 s! d1 z( m    } catch (Exception e) {
$ k& K: h/ r; x      e.printStackTrace (System.err);
    }
   
/ }: U& J+ Y( @+ ?* Y    syncUpdateOrder ();

+ x/ W+ p4 n5 \. H1 b% W    try {
% F, m8 L3 U9 O4 }! d      modelActions.createActionTo$message
        (heat, new Selector (heat.getClass (), "updateLattice", false));
5 v! ~0 Z7 Q: R( f% i4 D    } catch (Exception e) {
1 s% r* D6 q* r5 U6 |      System.err.println("Exception updateLattice: " + e.getMessage ());
    }
+ k  Y- D9 I  s  \' a        
: s2 ?' e  A" T    // Then we create a schedule that executes the
* a) _9 |6 D3 N" h6 C0 _    // 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
5 e; V7 O9 D. S    // modelActions ActionGroup at particular times.  This
    // schedule has a repeat interval of 1, it will loop every
    // time step.  The action is executed at time 0 relative to
6 e. y, T6 Z: x    // the beginning of the loop.
9 Z3 J, y' m, T0 z
    // This is a simple schedule, with only one action that is
7 o3 C7 P8 E7 Z4 s( P    // just repeated every time. See jmousetrap for more
    // complicated schedules.
  
    modelSchedule = new ScheduleImpl (getZone (), 1);
    modelSchedule.at$createAction (0, modelActions);
        
. w# e6 w. l' U4 c    return this;
6 ]& l# t0 _8 |, x# r) R7 ~" V5 c  }