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 () {
; I% |/ t) D) C6 h, |    super.buildActions();
   
    // Create the list of simulation actions. We put these in
- g) T& e& x- Q    // an action group, because we want these actions to be
- A( ?+ \9 j" d, O/ K6 E1 h2 E    // executed in a specific order, but these steps should
0 P, d: q% T; h) ?    // take no (simulated) time. The M(foo) means "The message
    // called <foo>". You can send a message To a particular
0 d! C+ d4 E1 b* U4 t- }) s    // object, or ForEach object in a collection.
        
    // Note we update the heatspace in two phases: first run
9 ~/ @! B8 J; _4 X" W    // diffusion, then run "updateWorld" to actually enact the
    // changes the heatbugs have made. The ordering here is
    // significant!
! A. n7 ]) J8 p        
2 N$ a7 d5 l# C. I& k0 u    // Note also, that with the additional
! G2 e7 B& |0 q( }    // `randomizeHeatbugUpdateOrder' Boolean flag we can
    // randomize the order in which the bugs actually run
1 A" n2 `( V* w% @3 M, r: d    // their step rule.  This has the effect of removing any
9 d! w5 z4 J( X8 S9 ^    // systematic bias in the iteration throught the heatbug
1 G! @2 h1 Y0 s2 o+ W: {    // list from timestep to timestep
  ]% ]# v- ~6 Z3 N: k) F        
    // By default, all `createActionForEach' modelActions have
    // a default order of `Sequential', which means that the
9 n' ]- K" u4 E$ A$ e1 ~; x    // order of iteration through the `heatbugList' will be
    // identical (assuming the list order is not changed
    // indirectly by some other process).
. T' \3 _: L1 M. k8 u9 ~   
# U0 z) `& }, t) a/ `0 `6 }$ f    modelActions = new ActionGroupImpl (getZone ());
! W0 U/ o0 B1 D- S. G
* r+ [: I+ I# g/ d6 ^% r/ `" C    try {
7 L  u0 N" k7 Z% {( f( ~      modelActions.createActionTo$message
        (heat, new Selector (heat.getClass (), "stepRule", false));
# F) h: H: ^: s1 r% [    } catch (Exception e) {
      System.err.println ("Exception stepRule: " + e.getMessage ());
    }
, l4 K, U0 e/ ?* y  z+ I
    try {
      Heatbug proto = (Heatbug) heatbugList.get (0);
+ u$ I+ N8 k+ F0 Q      Selector sel =
$ R1 ~) M, q9 e5 z, o) J+ Y) z        new Selector (proto.getClass (), "heatbugStep", false);
      actionForEach =
        modelActions.createFActionForEachHomogeneous$call
6 H0 A7 ?4 N- A' V        (heatbugList,
' p2 R, [  Z( p         new FCallImpl (this, proto, sel,
                        new FArgumentsImpl (this, sel)));
3 e. ]2 C1 `2 u5 N, ?: M    } catch (Exception e) {
      e.printStackTrace (System.err);
    }
* A0 m1 ?0 E0 J9 Q9 ]9 ~! g! g   
$ Q5 f! f2 T) g5 j  X* i    syncUpdateOrder ();
6 U) M: i0 B. d+ U# a2 ^+ R
    try {
      modelActions.createActionTo$message
7 e/ z% Y% ~% Y4 @4 r/ n# k        (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
; m9 E: B3 l) z: H    // modelActions. modelActions is an ActionGroup, by itself it
    // has no notion of time. In order to have it executed in
" [% {; m, u; G" m; G) V    // time, we create a Schedule that says to use the
    // modelActions ActionGroup at particular times.  This
- t5 Y0 c- f: p( a( Y3 B. x    // schedule has a repeat interval of 1, it will loop every
1 I) t' C7 [% \/ P    // time step.  The action is executed at time 0 relative to
    // the beginning of the loop.

- Y5 R/ e( k1 d$ x4 u: m# ~    // This is a simple schedule, with only one action that is
    // just repeated every time. See jmousetrap for more
" {! ^7 n4 r5 R6 \) j    // complicated schedules.
  
    modelSchedule = new ScheduleImpl (getZone (), 1);
    modelSchedule.at$createAction (0, modelActions);
        
0 d3 ~) `, @9 n" ?" T7 a# L5 {  B    return this;
( \1 W5 x3 ?0 A) K0 p  }