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：
4 g$ A- P/ S- D1 j: a/ n8 H
public Object buildActions () {
! x7 Y: }6 Q/ }    super.buildActions();
   
8 v7 x9 N. s: D( ^5 ]6 ?    // Create the list of simulation actions. We put these in
1 l& C' l1 f' }1 f9 C) n! I6 X5 M    // an action group, because we want these actions to be
) C! `% _4 |) i2 O0 t6 |    // executed in a specific order, but these steps should
    // take no (simulated) time. The M(foo) means "The message
- r9 K* w; w6 k3 P- ?* L    // called <foo>". You can send a message To a particular
    // object, or ForEach object in a collection.
  O, ~2 H; y$ o9 m3 r3 {        
    // Note we update the heatspace in two phases: first run
    // diffusion, then run "updateWorld" to actually enact the
6 g1 J% i1 ]2 b0 j    // changes the heatbugs have made. The ordering here is
    // significant!
        
* k9 D6 ~  Z2 g$ V2 q0 W% T% l    // Note also, that with the additional
    // `randomizeHeatbugUpdateOrder' Boolean flag we can
    // randomize the order in which the bugs actually run
/ s* \. ~( l0 E$ f) D0 P4 w    // their step rule.  This has the effect of removing any
; y) f, I" D) \" B( F4 \/ v    // systematic bias in the iteration throught the heatbug
: f5 n0 A7 Z8 t    // list from timestep to timestep
; R" Q8 k+ A" v        
    // By default, all `createActionForEach' modelActions have
    // a default order of `Sequential', which means that the
) N9 s) k' X- |, C2 a0 [9 T( y4 Q    // order of iteration through the `heatbugList' will be
    // identical (assuming the list order is not changed
    // indirectly by some other process).
5 q( D# S& g% F3 B9 u   
" z' C" ?; C3 V* u; N, Q    modelActions = new ActionGroupImpl (getZone ());

    try {
8 o" ^. n4 P; D3 ~7 B1 @      modelActions.createActionTo$message
        (heat, new Selector (heat.getClass (), "stepRule", false));
    } catch (Exception e) {
3 f# h; V/ \3 N. z      System.err.println ("Exception stepRule: " + e.getMessage ());
    }
3 l8 n& C8 ^. r: A# p: T
2 ]- X! ]  n+ n2 q, r    try {
% v$ A3 }) I5 g5 j      Heatbug proto = (Heatbug) heatbugList.get (0);
      Selector sel =
        new Selector (proto.getClass (), "heatbugStep", false);
* X0 W" U; a- K3 v3 ~; h      actionForEach =
        modelActions.createFActionForEachHomogeneous$call
5 ~# v8 g# b9 f8 m        (heatbugList,
+ H" C* C2 m+ Y- S( `- A6 t" I         new FCallImpl (this, proto, sel,
                        new FArgumentsImpl (this, sel)));
! `# L' U; y- O7 }; E+ M- Z    } catch (Exception e) {
      e.printStackTrace (System.err);
1 \1 D  Q  R) c8 [$ T/ r    }
   
    syncUpdateOrder ();
  {& l. |& h3 `6 m1 m
    try {
      modelActions.createActionTo$message
* Q# P0 m" \7 x8 m        (heat, new Selector (heat.getClass (), "updateLattice", false));
) p6 m/ C  F; \& t# A1 e3 E    } catch (Exception e) {
      System.err.println("Exception updateLattice: " + e.getMessage ());
, U3 x) t& @1 [# P  g    }
        
* ?5 s5 v" G" M0 K* U% U" e    // Then we create a schedule that executes the
! X: n! i1 `% ?9 {: a! i) C; \8 R    // modelActions. modelActions is an ActionGroup, by itself it
    // has no notion of time. In order to have it executed in
/ K% N& ~4 O0 a  _' E    // 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
    // the beginning of the loop.
0 ^5 U2 o- L0 f) w
, b1 I9 @! n; }: C- p$ Z$ v" @    // This is a simple schedule, with only one action that is
    // just repeated every time. See jmousetrap for more
8 u% c' K0 p* `; _% g    // complicated schedules.
: d* c# m7 a) J9 C1 Y  
1 U7 K" I& G  p: j; ?! @$ s    modelSchedule = new ScheduleImpl (getZone (), 1);
) [0 p$ M! u4 R8 Z* i, s, |    modelSchedule.at$createAction (0, modelActions);
4 ^2 w  E' J* O+ B- r" l( g        
    return this;
  }