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：
5 {1 S4 W$ ?/ K3 u0 w1 v
public Object buildActions () {
    super.buildActions();
   
2 _8 {1 H$ y7 f' e* K    // Create the list of simulation actions. We put these in
    // an action group, because we want these actions to be
- M4 g0 n; F9 [6 m0 {3 `3 H    // executed in a specific order, but these steps should
4 o& h- [3 E2 {& }# F! e    // 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.
0 b+ E# y% V' Y- [( B        
    // 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
8 \- e! A1 i2 G& a* o' ~5 y* G. @    // significant!
# K+ m% T) D% D, v1 D& d: ~) t        
# T9 U5 Z8 p, K, G2 l    // Note also, that with the additional
- [8 B# B  ]4 S! i0 B1 \5 T    // `randomizeHeatbugUpdateOrder' Boolean flag we can
    // randomize the order in which the bugs actually run
    // their step rule.  This has the effect of removing any
* K: o6 u  D0 N  v$ f" b6 {* h    // systematic bias in the iteration throught the heatbug
2 R2 ]5 g7 L, u% k    // list from timestep to timestep
        
7 u/ e# R7 B4 F$ U7 N9 q' Q    // By default, all `createActionForEach' modelActions have
' _& h; }2 Y6 X1 E: h    // a default order of `Sequential', which means that the
0 }% T4 K5 @7 H1 C3 N    // order of iteration through the `heatbugList' will be
    // identical (assuming the list order is not changed
    // indirectly by some other process).
/ r# j: N( x6 f# m5 Q$ y) Z% V   
% J$ Y' ]. x' l! M+ T( ~    modelActions = new ActionGroupImpl (getZone ());
' ]2 x$ V9 R) z. H9 S, |- c) o
    try {
4 ~4 a7 L- f. s6 q3 G& |- Q3 ~      modelActions.createActionTo$message
        (heat, new Selector (heat.getClass (), "stepRule", false));
9 D) t  h& S9 s9 |0 ^8 m    } catch (Exception e) {
      System.err.println ("Exception stepRule: " + e.getMessage ());
5 K0 o% y8 Y+ B- _. V, H    }
; }+ ]: d+ t1 c' b0 w# u# |& ]
" @$ o! G- J! \8 q% o5 f3 _    try {
: h8 @6 I/ m* t1 r& @      Heatbug proto = (Heatbug) heatbugList.get (0);
      Selector sel =
! c! \- k- P, Y$ z2 h5 u9 Z        new Selector (proto.getClass (), "heatbugStep", false);
. N4 z& c5 s, |# E3 y# e6 b      actionForEach =
1 Z' _% t: M+ z% y        modelActions.createFActionForEachHomogeneous$call
* A8 o* C& X5 y0 ~( [+ N3 g9 s        (heatbugList,
         new FCallImpl (this, proto, sel,
7 ~9 D4 y" c. m5 w8 ?* \                        new FArgumentsImpl (this, sel)));
    } catch (Exception e) {
      e.printStackTrace (System.err);
( q3 K+ e2 v8 Y$ m    }
   
: V3 A* ]* x: i3 b) a    syncUpdateOrder ();

    try {
      modelActions.createActionTo$message
        (heat, new Selector (heat.getClass (), "updateLattice", false));
- v" q( O) X3 V    } catch (Exception e) {
      System.err.println("Exception updateLattice: " + e.getMessage ());
    }
        
1 K' o7 b4 y+ u    // Then we create a schedule that executes the
    // modelActions. modelActions is an ActionGroup, by itself it
- a/ P% R5 b$ b$ \; b    // has no notion of time. In order to have it executed in
    // time, we create a Schedule that says to use the
. y1 ]0 h7 ~3 y- e    // 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
  }; L2 P8 K& L* v+ }2 l    // the beginning of the loop.

    // This is a simple schedule, with only one action that is
    // just repeated every time. See jmousetrap for more
( y; D8 \$ c  d2 ]& t$ R6 b    // complicated schedules.
) l- y" r' Q! ?! d1 z  
    modelSchedule = new ScheduleImpl (getZone (), 1);
8 A; P: e) N9 J( v    modelSchedule.at$createAction (0, modelActions);
        
    return this;
  }