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：

( s3 N* x; A2 L" d% v1 T; T9 Z public Object buildActions () {
    super.buildActions();
5 K+ f8 L& O4 b: j8 I3 P6 w/ r7 g   
8 w  L% h& }+ S4 ?1 ^$ J# h: ]' \6 D    // Create the list of simulation actions. We put these in
    // an action group, because we want these actions to be
  t6 g4 T; _7 V' t( q    // executed in a specific order, but these steps should
    // take no (simulated) time. The M(foo) means "The message
4 [3 V1 k; m* l% o& m    // called <foo>". You can send a message To a particular
! U/ e6 d, s9 ~7 A8 u3 N    // 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
    // changes the heatbugs have made. The ordering here is
) y- e8 s! @- y5 T- \    // significant!
        
" j* a  f5 i, K; K    // Note also, that with the additional
# D' C) Q2 g' z2 ^: L$ d6 [& u/ q    // `randomizeHeatbugUpdateOrder' Boolean flag we can
    // randomize the order in which the bugs actually run
    // their step rule.  This has the effect of removing any
) T( K7 C- m6 O0 D% x- G    // systematic bias in the iteration throught the heatbug
    // list from timestep to timestep
% m( ~: c' r) Q0 m        
    // By default, all `createActionForEach' modelActions have
    // 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
2 L1 z& `, o! a: z8 K, }. M    // indirectly by some other process).
   
: m# g! Y  W5 e1 e% Q4 j( d2 D    modelActions = new ActionGroupImpl (getZone ());
) R7 h8 v; I" x* S  q" o, [
, z1 ]" w! h3 s3 m' }0 F    try {
8 ^1 ?4 g8 u2 k$ b; E, g& `. V      modelActions.createActionTo$message
% D  U/ ]. H% w1 n        (heat, new Selector (heat.getClass (), "stepRule", false));
    } catch (Exception e) {
      System.err.println ("Exception stepRule: " + e.getMessage ());
9 \! O2 I4 A1 U) C6 B6 x2 E5 l    }

    try {
      Heatbug proto = (Heatbug) heatbugList.get (0);
      Selector sel =
        new Selector (proto.getClass (), "heatbugStep", false);
% Y( `* Y. s+ Q2 |7 X5 }; x5 x' N      actionForEach =
- A' q4 s, G) k& l! L- W" z0 i' T        modelActions.createFActionForEachHomogeneous$call
        (heatbugList,
         new FCallImpl (this, proto, sel,
& Y1 s- m  r& a2 b$ v9 R: Y                        new FArgumentsImpl (this, sel)));
3 h- V; @) o9 Q8 V    } catch (Exception e) {
      e.printStackTrace (System.err);
1 F5 V, c1 @, T6 ?5 Q  P; M- |    }
1 |/ e6 U- B( u, q1 b0 R" E  Q   
    syncUpdateOrder ();
+ C# u% N) l6 u" s
    try {
# D3 m0 T+ S9 k7 M      modelActions.createActionTo$message
* N! r3 s* l( j& }. t+ k        (heat, new Selector (heat.getClass (), "updateLattice", false));
9 ]3 Z& d6 r) G4 z) Q/ a) L    } catch (Exception e) {
7 s4 ^9 y7 Z6 X8 h7 l. P+ q      System.err.println("Exception updateLattice: " + e.getMessage ());
( g2 p2 r$ K2 Y, ?* |    }
        
    // Then we create a schedule that executes the
# {; u1 |: w" n  k% \7 n; B+ a    // 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
' w$ I3 t8 V! i0 C, O+ f    // schedule has a repeat interval of 1, it will loop every
( F  i3 ?. ^. v$ M# C4 R    // time step.  The action is executed at time 0 relative to
7 D4 [4 h  n5 `% P3 ]    // the beginning of the loop.
( X" }$ a5 _+ ~" A9 m+ }6 \3 R( \
    // This is a simple schedule, with only one action that is
6 j, @2 u* N3 x4 c    // just repeated every time. See jmousetrap for more
    // complicated schedules.
  
0 X: N$ P# L( ~    modelSchedule = new ScheduleImpl (getZone (), 1);
3 x0 N: b% w, V9 P8 X$ f    modelSchedule.at$createAction (0, modelActions);
        
2 Q% E* J% y4 L4 T9 A+ b6 \    return this;
  }