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 B! X% b( D6 n  y public Object buildActions () {
    super.buildActions();
   
/ l- g- {" d# b$ p' }    // Create the list of simulation actions. We put these in
8 m3 m$ @$ J: J6 {    // an action group, because we want these actions to be
) V; P, b5 Q- o) ]    // executed in a specific order, but these steps should
  b- I; j$ E3 H    // take no (simulated) time. The M(foo) means "The message
    // called <foo>". You can send a message To a particular
) S7 J% t5 r# [5 s- |" g    // object, or ForEach object in a collection.
2 E! g, k2 f5 A' g2 T5 c        
    // Note we update the heatspace in two phases: first run
1 ^: r1 E9 `( v8 Q% {    // diffusion, then run "updateWorld" to actually enact the
    // changes the heatbugs have made. The ordering here is
    // significant!
( Y+ x3 M3 V3 z1 m, b        
    // Note also, that with the additional
; Y' |" R9 `8 ?    // `randomizeHeatbugUpdateOrder' Boolean flag we can
7 n5 F0 K3 f4 d4 V: u    // randomize the order in which the bugs actually run
& ]7 V% W9 Q6 N4 p    // their step rule.  This has the effect of removing any
: D0 K: L+ Q: ]5 x9 g+ ^2 S! ?& J    // systematic bias in the iteration throught the heatbug
+ x; d0 F0 g5 r: s) Q; }    // list from timestep to timestep
        
7 d' i# X- K2 u; Y2 a" |    // By default, all `createActionForEach' modelActions have
% C5 q9 |" B5 m/ Y4 x* G    // a default order of `Sequential', which means that the
6 H% X! W/ [$ w% B    // order of iteration through the `heatbugList' will be
    // identical (assuming the list order is not changed
    // indirectly by some other process).
  ~$ y9 R0 R  m' f   
* y% f& q9 X" X+ S& u) Y# N' t    modelActions = new ActionGroupImpl (getZone ());
; }% K6 `3 Q/ j  N3 L- Z5 Z8 {
    try {
      modelActions.createActionTo$message
        (heat, new Selector (heat.getClass (), "stepRule", false));
    } catch (Exception e) {
      System.err.println ("Exception stepRule: " + e.getMessage ());
    }
  ], P8 q4 r  z; n7 _& @
    try {
! P; z! A- Y! a. N7 ?6 g, a( @6 Y      Heatbug proto = (Heatbug) heatbugList.get (0);
      Selector sel =
        new Selector (proto.getClass (), "heatbugStep", false);
" P+ `  O) Q7 ~' b      actionForEach =
4 A( h7 s/ B. w! t/ c/ O( A; Q" i        modelActions.createFActionForEachHomogeneous$call
) s* Z  @0 C' Y3 _7 B. K0 i* O9 E) E+ x        (heatbugList,
         new FCallImpl (this, proto, sel,
' e! X& ^4 I, |0 |                        new FArgumentsImpl (this, sel)));
    } catch (Exception e) {
' n2 r0 Z- Z5 {      e.printStackTrace (System.err);
    }
   
    syncUpdateOrder ();
) S# u! @! a! G7 M
! D' }6 I; p) d/ h  h+ D& H( p% R    try {
  V% }" u3 @, R2 h5 r0 @      modelActions.createActionTo$message
        (heat, new Selector (heat.getClass (), "updateLattice", false));
    } catch (Exception e) {
      System.err.println("Exception updateLattice: " + e.getMessage ());
' L4 w1 H5 D$ _5 J- K4 `2 \    }
        
    // Then we create a schedule that executes the
$ H, U$ Z  _5 ~. e1 u8 I. q$ _    // 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
9 Z+ J$ n4 v, }* n: G0 i) L    // modelActions ActionGroup at particular times.  This
1 @) z) n1 K, A: k$ I    // schedule has a repeat interval of 1, it will loop every
1 C7 C4 q% u( B7 M8 I    // time step.  The action is executed at time 0 relative to
    // the beginning of the loop.
" q; O8 ^8 \9 d7 ^! j
- \( l0 a' U! [    // This is a simple schedule, with only one action that is
6 F. b( f( q& L% i" J& D& U3 {8 o4 l% I. v    // just repeated every time. See jmousetrap for more
2 k, O4 M: Y8 d: C, ^    // complicated schedules.
2 h$ @! l$ p) [+ _  
    modelSchedule = new ScheduleImpl (getZone (), 1);
    modelSchedule.at$createAction (0, modelActions);
, g# g2 }! b, |* w        
    return this;
( _: J8 x$ G  \' i- Q$ v6 E* `& ~  }