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:
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public Object buildActions () {7 L& v( J" w2 z' Y
super.buildActions();$ W1 p; v' ?/ `
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// Create the list of simulation actions. We put these in8 U: p1 V* }- m) B* r- a, i
// an action group, because we want these actions to be: R, X5 G1 x% G1 W$ v
// executed in a specific order, but these steps should
' Y2 ?; |- O$ o! m: F/ l // take no (simulated) time. The M(foo) means "The message
* B7 T( l" G, a6 E# l // called <foo>". You can send a message To a particular# @- _6 p8 x1 ?1 B5 Y
// object, or ForEach object in a collection.
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// Note we update the heatspace in two phases: first run
: S: h) H/ z5 A6 R* N# R // diffusion, then run "updateWorld" to actually enact the
3 s) M& O/ z" J5 |' g7 _ // changes the heatbugs have made. The ordering here is
: m& P$ |/ s2 _# ~; `9 J+ k // significant!( ^0 f4 b3 X7 H5 K7 n) J! r
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// Note also, that with the additional2 J8 C" ^. a5 `2 ]
// `randomizeHeatbugUpdateOrder' Boolean flag we can
) O' Y4 e# `# ^ // randomize the order in which the bugs actually run# r$ Q+ {8 O* Q9 A1 [3 a8 M
// their step rule. This has the effect of removing any/ X" u# l9 H+ M) D# _2 g; x1 w
// systematic bias in the iteration throught the heatbug D$ o0 K' ? F, e8 ~( q* ?
// list from timestep to timestep
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// By default, all `createActionForEach' modelActions have2 J( D( h4 t1 a' ~; m
// a default order of `Sequential', which means that the
: y' ?$ Y8 S, _2 N // order of iteration through the `heatbugList' will be0 c1 h h/ C! x# G" c5 V
// identical (assuming the list order is not changed* ~. p6 D k5 z* U0 V [
// indirectly by some other process).
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modelActions = new ActionGroupImpl (getZone ());
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modelActions.createActionTo$message
! j3 \# w; c0 X. Z6 A. L" P (heat, new Selector (heat.getClass (), "stepRule", false));
: ?# J! }% n x } catch (Exception e) {
" S, I7 B3 y. b/ p# Z, c System.err.println ("Exception stepRule: " + e.getMessage ());
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9 G Z) t, v0 E3 k( [$ a try {
4 T: h# s& D8 }; ]8 g7 b Heatbug proto = (Heatbug) heatbugList.get (0);" v9 d7 d. u% [+ B# V' H; N4 H
Selector sel =
) H+ v7 W' S! f# e- e new Selector (proto.getClass (), "heatbugStep", false);& X, F& \9 X( U
actionForEach =
9 l. F+ I" M8 g5 M7 Q$ ? modelActions.createFActionForEachHomogeneous$call8 q$ x8 ]7 D1 g* y+ c) A
(heatbugList,
, j( ?0 A2 e! G, b, ] new FCallImpl (this, proto, sel,
7 K+ z1 H0 r- H- b" n8 S new FArgumentsImpl (this, sel)));
& v8 _- e% r+ o: a6 t( ~4 r } catch (Exception e) {9 B/ f& l9 V9 E0 K9 z
e.printStackTrace (System.err);
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' D' ~% D$ D7 j) E( D b- e syncUpdateOrder ();7 B- n) @, k G9 h: l' {% k
- S/ _+ O( O9 |9 R6 \, V8 |( f try {
! B( q u0 O: X0 R3 M2 } modelActions.createActionTo$message
3 v$ ~/ v4 L- ^9 D `3 } (heat, new Selector (heat.getClass (), "updateLattice", false));
5 d( q* b) a2 I9 D } catch (Exception e) {2 \, G3 Q" S2 M* @- m' G' l
System.err.println("Exception updateLattice: " + e.getMessage ());7 k$ T3 e7 _( J8 B% d2 ^4 ?
}
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// Then we create a schedule that executes the( j7 Y9 v. [. u. C7 Z1 ~
// modelActions. modelActions is an ActionGroup, by itself it
) m, h8 g6 m4 {: j4 p1 N4 K // has no notion of time. In order to have it executed in
/ w6 Q/ _2 T* p // time, we create a Schedule that says to use the0 b: ~8 M+ u6 z
// modelActions ActionGroup at particular times. This
+ ]1 L* i$ }* B* T9 k // schedule has a repeat interval of 1, it will loop every8 ~" v, g! ^7 m& m8 X0 y
// time step. The action is executed at time 0 relative to
% \9 f7 q* n6 Z2 P) S* _ // the beginning of the loop.3 u4 X6 o: R- R$ H& m Q
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// This is a simple schedule, with only one action that is" D) ~/ c# I8 f. s2 O$ Y( i/ a/ J8 w
// just repeated every time. See jmousetrap for more$ h0 Q$ p: Z [: h" B
// complicated schedules.# c9 N y5 {/ F6 g8 x
: j! [1 v- Q) p modelSchedule = new ScheduleImpl (getZone (), 1);, C# k. N5 m/ y
modelSchedule.at$createAction (0, modelActions);% }3 X+ M( K( d& z1 s0 B
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return this;- t2 n0 R* `3 D7 \) B
} |