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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+ j% v$ i3 c: C) F public Object buildActions () {7 S. _1 Q4 N f, t2 Z
super.buildActions();
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// Create the list of simulation actions. We put these in$ @9 g3 D* I1 @" Z- E# M
// an action group, because we want these actions to be
5 w* C: o5 h- ?% x, | // executed in a specific order, but these steps should b/ D( u' q3 y9 T y; r
// take no (simulated) time. The M(foo) means "The message
) M7 Y) b7 M5 V% J4 M& t // called <foo>". You can send a message To a particular+ K3 I0 g6 k6 ]# A
// object, or ForEach object in a collection." o' h" j0 w+ }* L" N- F" ?8 J/ p
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// Note we update the heatspace in two phases: first run0 o+ [3 |: @2 M+ ^ }# c/ X F! @9 r
// diffusion, then run "updateWorld" to actually enact the# S S# a7 J" S& U, a" m2 L3 R3 n
// changes the heatbugs have made. The ordering here is+ c( z- \ T5 t, ] @. L o
// significant!
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// Note also, that with the additional* `0 V/ |& Y1 T1 C' _' G8 E
// `randomizeHeatbugUpdateOrder' Boolean flag we can2 H$ g* S7 H0 m1 h: Z* {8 q# P( T
// randomize the order in which the bugs actually run
0 m: g! z8 _/ j2 K/ [ // their step rule. This has the effect of removing any
, ]! b3 u5 n% |4 f: j // systematic bias in the iteration throught the heatbug
8 M9 r2 D6 `/ V0 j& @ // list from timestep to timestep5 A: j- ]9 u8 W3 L7 j
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// By default, all `createActionForEach' modelActions have
# a( i8 `6 a# E2 {( G2 B# w0 J Z // a default order of `Sequential', which means that the
. ^; p/ J- {1 E# @: s( M. i // order of iteration through the `heatbugList' will be
+ e2 _- j, {0 j& G; W // identical (assuming the list order is not changed
1 r& t( W2 g1 H' a5 E2 D // indirectly by some other process).
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) R( f9 a& U$ G modelActions = new ActionGroupImpl (getZone ());
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try {
' w" j5 D9 F% x$ \ modelActions.createActionTo$message
! d+ D; I3 d. v/ I& z' o% m% Z4 n (heat, new Selector (heat.getClass (), "stepRule", false));
/ n) F& [: I& R: j } catch (Exception e) {
% P1 r) t; y8 o System.err.println ("Exception stepRule: " + e.getMessage ());* a3 a' `9 w2 W" {
}
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try {
2 X# u. o% P+ h% h- ^ j8 k: r Heatbug proto = (Heatbug) heatbugList.get (0);+ h$ w+ `! l" b ?5 I
Selector sel =
, |. r7 ?: O4 g" H( t( n6 B new Selector (proto.getClass (), "heatbugStep", false);" R% r1 z' Z6 Q, G" }/ ~# A7 _
actionForEach =$ T D- Y" x6 p$ o& d; q. W* s) }* F
modelActions.createFActionForEachHomogeneous$call4 W, q* `: t1 m/ K0 V# N9 d
(heatbugList,
7 O9 w' E' o& L! ~ new FCallImpl (this, proto, sel,$ t- {% j Y y k
new FArgumentsImpl (this, sel)));
* P* f9 ?5 j" E* M } catch (Exception e) {8 w1 A; `3 ], m
e.printStackTrace (System.err);
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syncUpdateOrder ();
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try {6 h3 U+ {* {8 V
modelActions.createActionTo$message ' s7 d& \; O+ p: H
(heat, new Selector (heat.getClass (), "updateLattice", false));
8 \. k$ H; i" \ } catch (Exception e) {
. A+ w' C# q# A; h s" `6 }+ k System.err.println("Exception updateLattice: " + e.getMessage ());3 D& k9 \, z0 |0 Z
}
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// Then we create a schedule that executes the( r9 [: ?# |& z6 W0 l5 H1 K( o! N3 A$ _
// modelActions. modelActions is an ActionGroup, by itself it
# U6 s$ d, G: Y, n // has no notion of time. In order to have it executed in/ k! e2 ]; S, k4 K5 J
// time, we create a Schedule that says to use the' Y( \: V8 w, F0 E6 q0 m( }
// modelActions ActionGroup at particular times. This
# |* `* W' s5 a7 X) x) H // schedule has a repeat interval of 1, it will loop every
8 Q+ o- |! { W8 k& q h // time step. The action is executed at time 0 relative to3 y2 h* @5 y- N: E7 M
// the beginning of the loop.
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' C+ E1 }/ V& E4 n& x7 F. u // This is a simple schedule, with only one action that is
# a8 ?( P- T# j4 |3 k+ i // just repeated every time. See jmousetrap for more" D6 r* k* M/ z0 N
// complicated schedules.; ]; H5 h* W! o( T$ g& C, r; S
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modelSchedule = new ScheduleImpl (getZone (), 1);+ N2 @2 A2 |5 X- ~2 f4 ?0 q0 [
modelSchedule.at$createAction (0, modelActions);# N. m& e9 ]. u, I: @
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return this;
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