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：
- F( w6 T6 C! }- |7 m% H
8 H- R3 g) @4 S) Y7 W public Object buildActions () {
    super.buildActions();
& `0 y4 ~1 w/ v2 W' q   
    // Create the list of simulation actions. We put these in
    // an action group, because we want these actions to be
) |, m; V% i' K3 c$ z    // executed in a specific order, but these steps should
8 `( D, ^7 z# |. j    // take no (simulated) time. The M(foo) means "The message
    // called <foo>". You can send a message To a particular
6 m' G4 P, v1 H: y" g: R" n    // object, or ForEach object in a collection.
        
- F8 k, p- d" T+ b' ]1 C: [    // Note we update the heatspace in two phases: first run
    // diffusion, then run "updateWorld" to actually enact the
" s. E: V. @% w* C- ]    // changes the heatbugs have made. The ordering here is
8 Q& G: B4 m9 R4 s    // significant!
1 [4 P" `1 g  ^        
) H" Y! U, z8 H% e3 }    // Note also, that with the additional
    // `randomizeHeatbugUpdateOrder' Boolean flag we can
: P3 Z7 F1 Y' \/ l# p    // randomize the order in which the bugs actually run
5 ~$ a/ k; |9 _, X    // their step rule.  This has the effect of removing any
- c( z! t  W3 c2 J) @/ ]! K    // systematic bias in the iteration throught the heatbug
    // list from timestep to timestep
        
    // By default, all `createActionForEach' modelActions have
- t2 C" U" {# W) ?    // 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
* L0 U8 q  x( M    // indirectly by some other process).
   
- D  ?0 k0 h8 Y0 I    modelActions = new ActionGroupImpl (getZone ());
! u- ]$ f) e1 v. U, I7 T! Z
$ B; k5 Z) e4 x, u3 A1 f6 g% t    try {
% T! i& L- C# A' p% f      modelActions.createActionTo$message
$ I  A" H, c/ D# w' Y+ J2 _  V- W        (heat, new Selector (heat.getClass (), "stepRule", false));
    } catch (Exception e) {
- b* C! d9 m+ g2 |$ y      System.err.println ("Exception stepRule: " + e.getMessage ());
    }
: a8 n% l  K+ ?9 A8 R
  q4 L* Z& ~9 b    try {
      Heatbug proto = (Heatbug) heatbugList.get (0);
      Selector sel =
        new Selector (proto.getClass (), "heatbugStep", false);
+ ?  |+ w* Z0 c      actionForEach =
6 J2 |  w* u0 @/ G- J- Z8 i        modelActions.createFActionForEachHomogeneous$call
1 u& C1 y  p1 T# m  f        (heatbugList,
/ s+ ^5 B2 \  E* A9 _: o         new FCallImpl (this, proto, sel,
/ o. Z4 l- C" s/ Y2 \$ B* a$ Q4 X                        new FArgumentsImpl (this, sel)));
4 W9 E. W5 U6 G8 ^    } catch (Exception e) {
      e.printStackTrace (System.err);
    }
   
    syncUpdateOrder ();
# t5 C# R- H$ @# t" b
- R9 `: z, n. U$ k5 n$ B" l, K4 C    try {
      modelActions.createActionTo$message
        (heat, new Selector (heat.getClass (), "updateLattice", false));
    } catch (Exception e) {
1 V/ L& R0 }7 t      System.err.println("Exception updateLattice: " + e.getMessage ());
. D$ X. r5 O2 ~( B/ I. h    }
1 V' D0 A! L4 u/ P4 z2 `        
    // Then we create a schedule that executes the
6 R$ L; S+ m# u' p* d    // 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
    // modelActions ActionGroup at particular times.  This
7 p; t5 ]9 X. v$ P0 a2 J. O    // schedule has a repeat interval of 1, it will loop every
    // time step.  The action is executed at time 0 relative to
( F' o! @2 J6 @4 }, ]" a    // the beginning of the loop.

3 c+ ^$ B3 e4 w* v    // This is a simple schedule, with only one action that is
    // just repeated every time. See jmousetrap for more
    // complicated schedules.
( c' A! t$ V! }5 N& Z  
    modelSchedule = new ScheduleImpl (getZone (), 1);
    modelSchedule.at$createAction (0, modelActions);
        
$ J4 h' m+ l5 ^9 A% G/ P! _    return this;
4 `2 o3 u% E9 G  O  }