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:, T3 u5 t( l4 R: P
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public Object buildActions () {7 ?9 c, g9 |% @8 `7 ^# ?* c2 s/ ]
super.buildActions();, G7 Y, v: b/ H, k% y4 e
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// Create the list of simulation actions. We put these in: |1 ?$ l) a- K" a5 q
// an action group, because we want these actions to be
8 w3 e; x( l+ \7 V // executed in a specific order, but these steps should
8 Y6 I9 W; D) L# |: ^% H // take no (simulated) time. The M(foo) means "The message# I% }5 Z2 f3 E; Z. f3 U
// called <foo>". You can send a message To a particular* u+ [7 L7 ~3 M
// object, or ForEach object in a collection.
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7 ^5 ]0 ]7 N' p, t$ _8 U4 Q# g' r( N // Note we update the heatspace in two phases: first run
0 j1 F3 C& s; Q4 E% e+ P0 ^ // diffusion, then run "updateWorld" to actually enact the& K0 d( Z6 U& b- r0 W/ `2 N. V$ S
// changes the heatbugs have made. The ordering here is: d8 K" @8 k( M8 z
// significant!) z1 |4 m8 g& m; ~5 ~& W ]" ^
& t8 ?" {# |' {- P7 l // Note also, that with the additional
) o7 {* n. j! p) C, J) x) O // `randomizeHeatbugUpdateOrder' Boolean flag we can
1 @! j6 }$ k: q* n8 S( T // randomize the order in which the bugs actually run
+ S* b; @7 t4 r9 E // their step rule. This has the effect of removing any0 b7 l# C# b9 j! d
// systematic bias in the iteration throught the heatbug' k( x3 Q4 v! `
// list from timestep to timestep
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// By default, all `createActionForEach' modelActions have
5 E8 ^0 A0 t, T& w& V- J // a default order of `Sequential', which means that the
0 z3 K. C7 y" @ // order of iteration through the `heatbugList' will be7 m9 r3 f t0 s& I) n, t
// identical (assuming the list order is not changed
1 x4 H2 s! g: \3 | // indirectly by some other process).
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modelActions = new ActionGroupImpl (getZone ());
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try {* @3 @% j, h# l% t+ Z# A3 `) ^
modelActions.createActionTo$message5 Q( e0 n7 d' z6 _1 H6 n) x
(heat, new Selector (heat.getClass (), "stepRule", false));: I! h1 z3 I- @/ W" I1 P
} catch (Exception e) {
9 l! d; L: D$ g+ q: b9 o l% C System.err.println ("Exception stepRule: " + e.getMessage ());+ [1 b. y' J5 w, x
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try {, c1 g: F3 {3 `1 A2 G+ I! b+ V) l
Heatbug proto = (Heatbug) heatbugList.get (0);" q% ]9 A8 j# E' O6 `. v/ s- Y
Selector sel = ( s. Q0 \: y: L
new Selector (proto.getClass (), "heatbugStep", false);
/ I0 _ {+ A: x" ^$ l5 V actionForEach =% v4 Q3 g2 c2 O( ]
modelActions.createFActionForEachHomogeneous$call
7 j7 y0 m8 B! r2 }- v4 b (heatbugList,, ^# \+ _2 ^3 H) C% M
new FCallImpl (this, proto, sel,& @+ J( C+ t7 B, o( a' Y4 L
new FArgumentsImpl (this, sel)));& R4 F2 k- W6 k+ @% ~- |( c
} catch (Exception e) {
+ F y* V7 m% a; l/ y e.printStackTrace (System.err);; v, V; O4 k% s+ r
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syncUpdateOrder ();
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modelActions.createActionTo$message
0 W; i5 p9 K& b$ }4 [ (heat, new Selector (heat.getClass (), "updateLattice", false));
" } |3 @. i+ _# P% F. `' r } catch (Exception e) {
; i }: _, X$ Z8 G C) J8 O System.err.println("Exception updateLattice: " + e.getMessage ());
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' ]+ t% T2 A( G // Then we create a schedule that executes the
" L) G% C1 K8 l( x' L7 d // modelActions. modelActions is an ActionGroup, by itself it
" k) `; C! F! w) _( l // has no notion of time. In order to have it executed in
" M# G) a. d. l" L // time, we create a Schedule that says to use the/ f& {7 @6 q2 D- x0 U# j: _
// modelActions ActionGroup at particular times. This
) ~0 u0 h6 k ~ // schedule has a repeat interval of 1, it will loop every0 [0 d6 F' T) C* m2 }6 l4 a
// time step. The action is executed at time 0 relative to
' }$ d5 M+ l& Z$ A, R" I2 o // the beginning of the loop.
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// This is a simple schedule, with only one action that is
r) Y3 @* w. K& @7 v$ B) L // just repeated every time. See jmousetrap for more! h: g! q# u& B+ _
// complicated schedules.
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, ^( f5 `. K8 H7 d6 D modelSchedule = new ScheduleImpl (getZone (), 1);" }, c1 A0 d3 N9 }' C' H
modelSchedule.at$createAction (0, modelActions);
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return this;
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