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：
# `. D# k; i2 J1 A% w# `
public Object buildActions () {
    super.buildActions();
   
    // Create the list of simulation actions. We put these in
    // an action group, because we want these actions to be
( V. G( f( z* z2 C2 r+ ?  h: V    // executed in a specific order, but these steps should
3 ?1 X7 r: I' q! k    // take no (simulated) time. The M(foo) means "The message
6 g/ J! U- s8 ~6 k# w8 R' |/ M/ C    // called <foo>". You can send a message To a particular
% p8 A# D: b7 Y) Q" O( |    // object, or ForEach object in a collection.
        
6 x: M6 R) g- @    // Note we update the heatspace in two phases: first run
: c% ^2 m% Z) g" M7 n( @$ O) Q    // diffusion, then run "updateWorld" to actually enact the
7 |* {: v1 ]1 c+ Z$ V; Z, ^" F- {    // changes the heatbugs have made. The ordering here is
    // significant!
        
4 }  F1 e  S) H, Y) F    // Note also, that with the additional
. ]7 N  a) z) @6 T0 I0 {    // `randomizeHeatbugUpdateOrder' Boolean flag we can
    // 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
, F/ B$ E& P6 q+ `& Z! a, L  a2 K5 g    // list from timestep to timestep
        
    // By default, all `createActionForEach' modelActions have
% t, W0 ~9 m7 I) V' }    // a default order of `Sequential', which means that the
    // order of iteration through the `heatbugList' will be
$ S. }0 F% J: f" V5 V4 @5 i    // identical (assuming the list order is not changed
0 u5 r! P/ R( g    // indirectly by some other process).
3 U' Y' a7 Y5 z   
# i( J: m# c  a/ H    modelActions = new ActionGroupImpl (getZone ());

& g) ?% z7 I: {; Q; ~) D    try {
      modelActions.createActionTo$message
) _: ~. h* O  m9 b        (heat, new Selector (heat.getClass (), "stepRule", false));
    } catch (Exception e) {
! L) Z2 e. ]' @0 \      System.err.println ("Exception stepRule: " + e.getMessage ());
; I5 [7 k: Q4 M! s    }
: @* H- d" \3 F
& }8 u7 k- y- M# b) O) p    try {
      Heatbug proto = (Heatbug) heatbugList.get (0);
      Selector sel =
# ?# k( m- C* D+ i( y        new Selector (proto.getClass (), "heatbugStep", false);
      actionForEach =
        modelActions.createFActionForEachHomogeneous$call
1 h8 v1 R/ E( U) K$ C2 A        (heatbugList,
( v1 e  t2 r' |1 i. m         new FCallImpl (this, proto, sel,
                        new FArgumentsImpl (this, sel)));
    } catch (Exception e) {
      e.printStackTrace (System.err);
: [  S; _$ |8 I    }
9 _/ p# O8 i( [% G5 ~7 i" e- ^   
    syncUpdateOrder ();
' a6 A# L, m3 i5 T0 _
    try {
      modelActions.createActionTo$message
2 H7 w7 R0 A5 A/ F5 [1 [        (heat, new Selector (heat.getClass (), "updateLattice", false));
    } catch (Exception e) {
) j) m, l+ Y, V/ P# a. e# ]9 _      System.err.println("Exception updateLattice: " + e.getMessage ());
! S# L9 A- R8 |# r. ]* r    }
        
, Z1 ?8 c0 @6 [, L3 S9 F    // Then we create a schedule that executes the
- {0 q8 o7 A6 ^9 [    // modelActions. modelActions is an ActionGroup, by itself it
    // has no notion of time. In order to have it executed in
/ g3 T3 w8 b" u0 o* v7 L    // time, we create a Schedule that says to use the
    // modelActions ActionGroup at particular times.  This
2 ?. I9 P" v& b. R    // schedule has a repeat interval of 1, it will loop every
2 P; ^% R: g' ~2 H7 f7 ?% q8 Q* a    // time step.  The action is executed at time 0 relative to
6 Y1 s8 g: {+ R    // the beginning of the loop.
5 U1 b2 l7 v$ ^/ M
    // This is a simple schedule, with only one action that is
; f% @" s* A* m2 m+ g1 S    // just repeated every time. See jmousetrap for more
9 B# F+ _6 i3 W1 X9 A8 I    // complicated schedules.
  
! B) j) Y- ~' i1 f/ ~/ q    modelSchedule = new ScheduleImpl (getZone (), 1);
    modelSchedule.at$createAction (0, modelActions);
        
9 y8 c  E6 J6 j. F9 Q5 J    return this;
  }