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 () {
0 {/ y1 b' h1 [5 P, i4 e    super.buildActions();
   
    // Create the list of simulation actions. We put these in
    // an action group, because we want these actions to be
    // executed in a specific order, but these steps should
    // take no (simulated) time. The M(foo) means "The message
, ]9 s4 Z- Y1 o) t* B: ~/ ^    // called <foo>". You can send a message To a particular
+ D2 a' \9 `" w' H& j5 d( J( ?    // object, or ForEach object in a collection.
6 B% u' b: u3 v8 j- f        
    // Note we update the heatspace in two phases: first run
    // diffusion, then run "updateWorld" to actually enact the
4 e5 F/ v2 S- D& d+ {' i    // changes the heatbugs have made. The ordering here is
    // significant!
        
- e. a0 K2 a5 N( @7 [8 n    // Note also, that with the additional
    // `randomizeHeatbugUpdateOrder' Boolean flag we can
5 q4 H- C: I9 x2 u7 }8 S    // randomize the order in which the bugs actually run
    // their step rule.  This has the effect of removing any
    // systematic bias in the iteration throught the heatbug
& c" D% L% T# D4 @$ t    // list from timestep to timestep
/ K6 m* L- e5 P4 T4 L        
4 {1 S! M( n, V    // By default, all `createActionForEach' modelActions have
: K. R) z5 S4 \    // a default order of `Sequential', which means that the
3 i2 g  j7 B. y9 l8 n" u& }    // order of iteration through the `heatbugList' will be
    // identical (assuming the list order is not changed
( @+ j3 C& v0 N1 Y9 ?+ t  `    // indirectly by some other process).
   
2 t0 ~, a$ N; `* L9 A  s    modelActions = new ActionGroupImpl (getZone ());
) e4 J7 R) s5 T. |# V" Y
) Z9 m  k/ j' }7 T    try {
5 y3 m3 j. B+ \      modelActions.createActionTo$message
        (heat, new Selector (heat.getClass (), "stepRule", false));
    } catch (Exception e) {
      System.err.println ("Exception stepRule: " + e.getMessage ());
6 ]& K6 i: r8 n/ F/ ]' |8 Z    }
) @4 B' |. b5 {0 X* J% o+ ?
    try {
      Heatbug proto = (Heatbug) heatbugList.get (0);
' V9 k* ]+ h1 V1 s  O      Selector sel =
        new Selector (proto.getClass (), "heatbugStep", false);
3 n: y6 o( L0 A  H) {      actionForEach =
        modelActions.createFActionForEachHomogeneous$call
        (heatbugList,
+ @+ k  z( R# n         new FCallImpl (this, proto, sel,
                        new FArgumentsImpl (this, sel)));
2 T! v' E6 A) D1 e! ?    } catch (Exception e) {
  _! U- i8 X+ k1 O2 \0 g      e.printStackTrace (System.err);
: o  f; b$ ~, e1 E. _    }
   
    syncUpdateOrder ();

    try {
      modelActions.createActionTo$message
6 o5 N! T* C# X( k9 `; |6 }( s        (heat, new Selector (heat.getClass (), "updateLattice", false));
    } catch (Exception e) {
      System.err.println("Exception updateLattice: " + e.getMessage ());
    }
( q: L/ v* \0 l0 W' A% s        
, @. c. }0 a* {& H8 h' M    // Then we create a schedule that executes the
    // modelActions. modelActions is an ActionGroup, by itself it
: S/ F+ i5 b1 U: {    // 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
    // schedule has a repeat interval of 1, it will loop every
5 H. w5 C4 i) O6 ^' c7 V    // time step.  The action is executed at time 0 relative to
    // the beginning of the loop.

9 u2 o" k# X  j5 P  [* L& i1 b" R    // This is a simple schedule, with only one action that is
    // just repeated every time. See jmousetrap for more
    // complicated schedules.
  
- u! W' [. w2 ]    modelSchedule = new ScheduleImpl (getZone (), 1);
6 z3 v7 w( M+ E" Y; C4 ^    modelSchedule.at$createAction (0, modelActions);
5 L3 I) g7 A4 [! Q* ]        
+ W# t1 V5 x; e" A, }    return this;
  }