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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8 S- U) h! H& }* p& t public Object buildActions () {
    super.buildActions();
) T9 ^9 r* a, `   
$ B2 G. }- [+ L$ O5 d    // Create the list of simulation actions. We put these in
. \  o, v$ `+ l* G  Q  M4 A, [) u    // 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, Q7 B' \6 m$ C. r( z        
    // 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
6 X% B: R9 y8 F4 F    // significant!
2 E, ?/ n/ r0 N( S+ b( q        
" ~8 K& r9 S. W    // Note also, that with the additional
/ r, P7 Y) f: [1 Y; t3 W* R# v    // `randomizeHeatbugUpdateOrder' Boolean flag we can
& I8 \7 u* W4 H, ?, n8 {& i  Z    // 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
) J4 F' q- N2 B' b' \- `6 M; u    // list from timestep to timestep
+ ?7 v! H+ @  i) p$ O' F        
    // By default, all `createActionForEach' modelActions have
    // a default order of `Sequential', which means that the
: ^, p8 ^6 h' }6 W" O    // order of iteration through the `heatbugList' will be
( l7 c' L1 ?5 C; ]+ h    // identical (assuming the list order is not changed
4 Z0 q/ v: l, r1 ~  R+ ^    // indirectly by some other process).
5 x, N0 B$ [3 V2 h% H   
1 `+ V3 K5 B9 V* a    modelActions = new ActionGroupImpl (getZone ());

    try {
      modelActions.createActionTo$message
        (heat, new Selector (heat.getClass (), "stepRule", false));
8 W+ A+ f/ G$ z! t+ m/ l    } catch (Exception e) {
      System.err.println ("Exception stepRule: " + e.getMessage ());
    }
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% r4 G9 u! o, O: P0 }    try {
      Heatbug proto = (Heatbug) heatbugList.get (0);
' I% v1 N$ h1 X. g0 l      Selector sel =
        new Selector (proto.getClass (), "heatbugStep", false);
& n: U( G5 j1 B' s6 w6 O; S      actionForEach =
# r& _8 G. L( k/ O! n! }, j8 h        modelActions.createFActionForEachHomogeneous$call
, |& A3 J6 u$ \6 |8 u& t* g        (heatbugList,
, q! X. _" u  R) G: D+ P# ?' v% [7 c         new FCallImpl (this, proto, sel,
                        new FArgumentsImpl (this, sel)));
    } catch (Exception e) {
1 m: S7 ~$ T3 {! V2 Q7 Z7 R      e.printStackTrace (System.err);
    }
: ~# f0 r, P! _- D( e; y   
    syncUpdateOrder ();

    try {
9 X+ q) U/ ^: X1 s- s$ [) y" m      modelActions.createActionTo$message
        (heat, new Selector (heat.getClass (), "updateLattice", false));
    } catch (Exception e) {
      System.err.println("Exception updateLattice: " + e.getMessage ());
    }
        
    // Then we create a schedule that executes the
; [7 j6 G2 K! Q$ S. `  F    // modelActions. modelActions is an ActionGroup, by itself it
$ T1 c5 c" z# C  G    // has no notion of time. In order to have it executed in
0 X8 g) c9 ]# I* m, R% C    // 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
    // time step.  The action is executed at time 0 relative to
: i) ^6 Y5 a( L  L9 K    // the beginning of the loop.

1 i" e2 g7 r' r- d) L. N9 L    // This is a simple schedule, with only one action that is
    // just repeated every time. See jmousetrap for more
. i. u: Q" x# ^6 _    // complicated schedules.
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    modelSchedule = new ScheduleImpl (getZone (), 1);
    modelSchedule.at$createAction (0, modelActions);
        
    return this;
6 Z) `4 L" g: `" a, g5 g$ `  }