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：
: L' E3 i! \! r7 ^
public Object buildActions () {
9 u. Z* Y5 m) r. i    super.buildActions();
   
. e. Z+ g# i  F: q' f9 V3 K' m    // Create the list of simulation actions. We put these in
- M0 L: A7 v# a# g    // an action group, because we want these actions to be
# }2 s. o. y. V! F# Q, e, g" O    // executed in a specific order, but these steps should
0 [& {; s9 c! e* O5 z. Q6 @    // 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.
        
    // Note we update the heatspace in two phases: first run
    // diffusion, then run "updateWorld" to actually enact the
( N# q' o5 P* ^8 G3 e    // changes the heatbugs have made. The ordering here is
4 h% a7 P; c" Y    // significant!
        
6 H9 T4 x, e  A+ e: L! H5 w4 J    // Note also, that with the additional
2 b/ E0 ^: b$ w  t- n, V    // `randomizeHeatbugUpdateOrder' Boolean flag we can
    // randomize the order in which the bugs actually run
4 t  o2 E; f" N" v0 M+ z    // their step rule.  This has the effect of removing any
    // systematic bias in the iteration throught the heatbug
1 j; r, P! h, ~6 s    // list from timestep to timestep
        
7 m7 k2 C2 g& i; U% S    // By default, all `createActionForEach' modelActions have
4 {+ s% Z$ y, {  u    // a default order of `Sequential', which means that the
    // order of iteration through the `heatbugList' will be
1 F! A6 ?' u; Z& J& \' c0 z) [( u    // identical (assuming the list order is not changed
    // indirectly by some other process).
   
    modelActions = new ActionGroupImpl (getZone ());
; Q% O# w% U$ `# J) g5 S# }
    try {
7 e7 d/ W! F9 T/ m6 k3 V      modelActions.createActionTo$message
        (heat, new Selector (heat.getClass (), "stepRule", false));
) B- V2 M1 g" J$ i, {    } catch (Exception e) {
      System.err.println ("Exception stepRule: " + e.getMessage ());
    }

    try {
      Heatbug proto = (Heatbug) heatbugList.get (0);
      Selector sel =
9 h- Q3 o2 E# B* ~( |        new Selector (proto.getClass (), "heatbugStep", false);
      actionForEach =
4 e0 s/ ^7 `; @$ E        modelActions.createFActionForEachHomogeneous$call
        (heatbugList,
! o5 g' _  v9 S$ }3 X6 w7 Y' J         new FCallImpl (this, proto, sel,
- Y! p3 j* L$ o2 P                        new FArgumentsImpl (this, sel)));
8 T; X) n2 T; a. p3 Y    } catch (Exception e) {
      e.printStackTrace (System.err);
( j3 b  {3 ^5 s, N" Y    }
   
    syncUpdateOrder ();
) ^! n, _) W- Z) ]; x2 S( l  m- \- P
    try {
      modelActions.createActionTo$message
$ y' m& u/ g$ K3 H. H* b        (heat, new Selector (heat.getClass (), "updateLattice", false));
    } catch (Exception e) {
      System.err.println("Exception updateLattice: " + e.getMessage ());
3 j0 ?! a1 j6 s, f' s* `7 q4 F    }
        
, Y& g, }' s' ?3 U: V$ e    // Then we create a schedule that executes the
9 r) G4 i. Q8 G/ E) v0 x# d    // modelActions. modelActions is an ActionGroup, by itself it
& }+ G" K: S1 `    // 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
1 y0 O$ U6 \1 u    // schedule has a repeat interval of 1, it will loop every
2 ^5 Z/ Z% y: W) s    // time step.  The action is executed at time 0 relative to
8 t  J/ `4 c/ h8 b# ^    // the beginning of the loop.
  k& M' b7 [" v3 a4 i# R# r& }
6 E7 L$ [5 d, W# a& e. B7 ?    // This is a simple schedule, with only one action that is
    // just repeated every time. See jmousetrap for more
, T% X, \3 F* T  [  r    // complicated schedules.
5 J/ a6 q+ w- i5 j  
  E$ Y$ }: {* [0 E    modelSchedule = new ScheduleImpl (getZone (), 1);
4 \3 Q- ^9 k2 c9 J8 o    modelSchedule.at$createAction (0, modelActions);
( D; ?% r. Y+ g7 B        
    return this;
) E' h* ^# h) z5 r1 T6 H  }