设为首页收藏本站

最大的系统仿真与系统优化公益交流社区

 找回密码
 注册

QQ登录

只需一步,快速开始

查看: 12250|回复: 0

[求助] 问jheatbugs-2001-03-28中某些代码

[复制链接]
发表于 2008-5-25 02:15:22 | 显示全部楼层 |阅读模式
5仿真币
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:
# R5 L+ l# B  f1 v- {3 i$ u! `0 B2 y3 [/ r, I5 `
public Object buildActions () {
$ m# ]# U# `# q9 s! N  [    super.buildActions();
0 q5 A' f  M' w3 j    6 C: @9 t  K" Y/ E, T3 f& X
    // Create the list of simulation actions. We put these in4 Q& Z& B6 _4 T9 V4 E
    // an action group, because we want these actions to be
6 r) U; }6 X5 N8 }' Q; y. c6 i: i    // executed in a specific order, but these steps should
/ h" f7 I6 q( R4 J( p    // take no (simulated) time. The M(foo) means "The message
5 u: O( P# M* b. r% N    // called <foo>". You can send a message To a particular  }: m4 `; [$ R
    // object, or ForEach object in a collection.& L1 N% g0 k; o* U& p/ h
        
" k( q/ g# h  J: R9 I    // Note we update the heatspace in two phases: first run
. Y* d) q, e" q2 A' f3 D, q    // diffusion, then run "updateWorld" to actually enact the
& T% I# q, H: F  e$ ]+ C, ~    // changes the heatbugs have made. The ordering here is5 K1 r  \( K* `+ A
    // significant!3 w9 Y  f3 p6 R# _6 \' K
        # B% C! |: ^' v* M& c( h6 B1 B
    // Note also, that with the additional4 x, ^0 M+ J1 t1 Q' d
    // `randomizeHeatbugUpdateOrder' Boolean flag we can; Y- c: R9 d0 w
    // randomize the order in which the bugs actually run2 x6 L* s- I& _1 E9 L" s6 ~8 M  j% g
    // their step rule.  This has the effect of removing any
3 `3 v5 y0 v- ~$ z+ Q" y    // systematic bias in the iteration throught the heatbug+ Z! ?9 _+ w. L, m
    // list from timestep to timestep
; }, c- d* J: v  Z        6 o) q6 L4 J; o( R( ]: r
    // By default, all `createActionForEach' modelActions have
$ y1 h" E. Z, g$ v; l- Q6 n& o    // a default order of `Sequential', which means that the
, M9 q% H: D. B2 P    // order of iteration through the `heatbugList' will be9 F+ u  A% @$ i2 p' Q# ~
    // identical (assuming the list order is not changed0 ^6 i7 k9 k9 Z! Y& ?6 `4 m( O
    // indirectly by some other process).5 F$ V& {& Y/ a! Q, a* \
   
( y8 g, }6 z+ w2 Z% j8 @$ ?+ o    modelActions = new ActionGroupImpl (getZone ());. o$ U8 b9 s5 _0 ]
5 l  Q, p& V5 t" q) X. Q
    try {
8 n. p+ h) n1 C3 s      modelActions.createActionTo$message) y& C+ J% m3 `& L/ [
        (heat, new Selector (heat.getClass (), "stepRule", false));
6 n/ ~& g& B! R- k/ Y+ g    } catch (Exception e) {3 i2 D" y/ [" R  F
      System.err.println ("Exception stepRule: " + e.getMessage ());
! Y- `& |0 ^1 j& U( i7 r& x    }$ E  z7 Y6 n; ~* a$ m$ S

  V( A/ C& ^0 J- _( r7 n6 y    try {' g. M; a$ s. E& P
      Heatbug proto = (Heatbug) heatbugList.get (0);* e4 m0 F7 d/ |+ M
      Selector sel =
- A* ]" ]# C) s9 r  S        new Selector (proto.getClass (), "heatbugStep", false);8 X6 _& S/ M. ~! U. W4 T: U- T- i
      actionForEach =
- Q6 ~& a/ `* n2 t- Y5 |/ f$ z        modelActions.createFActionForEachHomogeneous$call
- m) e7 I  r7 y, u# ?: p( B        (heatbugList,) u1 \8 |  j* L  Q! w1 h$ a9 p# i
         new FCallImpl (this, proto, sel,
# I  o, ?0 c3 s8 X                        new FArgumentsImpl (this, sel)));/ y0 \( U4 w# ^, {+ R, e( z
    } catch (Exception e) {
2 \& k2 `% D6 X9 {      e.printStackTrace (System.err);0 V% O! B; |7 r$ N' p8 O3 B, @
    }
0 g# n8 V3 \) X5 [   
6 J! `" l  ^! |- I; K8 K' u$ I    syncUpdateOrder ();. X% l# V8 B* W& T

# |9 x/ ^( g7 X' q    try {
2 J9 c+ m- D, |* s7 ~      modelActions.createActionTo$message 8 m* G1 c& F9 H' J  ]) q
        (heat, new Selector (heat.getClass (), "updateLattice", false));
) q7 T8 n/ C# n9 w    } catch (Exception e) {5 d* O& _, b1 e' C; ^! J
      System.err.println("Exception updateLattice: " + e.getMessage ());, G/ C1 i" ]! [% y. H
    }
+ E5 F! A5 |/ [- D5 v        
! {& y2 e' R3 K4 n    // Then we create a schedule that executes the
5 c& P( Q- e( U1 x" h    // modelActions. modelActions is an ActionGroup, by itself it2 W+ B, n0 z+ P% L# r
    // has no notion of time. In order to have it executed in; r: g5 V4 j: ~
    // time, we create a Schedule that says to use the
3 Y: s5 P* b4 D3 N% Z& I    // modelActions ActionGroup at particular times.  This$ _; C# L9 q/ V
    // schedule has a repeat interval of 1, it will loop every
% F7 M# [, U* R! k  q5 E    // time step.  The action is executed at time 0 relative to) f* ]& y( f! _+ o0 n
    // the beginning of the loop.$ a9 Z% |5 h1 J8 Z

# s( j. n3 A) N2 s8 @    // This is a simple schedule, with only one action that is
3 e  y" c0 o; p3 @! _    // just repeated every time. See jmousetrap for more- Z/ `5 T, o; H2 V
    // complicated schedules.
3 g3 B5 T) {+ a! h& `) H2 M8 S  . j% B0 ]. q8 ?) H. j9 `
    modelSchedule = new ScheduleImpl (getZone (), 1);1 j% [  u; L' m8 X
    modelSchedule.at$createAction (0, modelActions);9 d+ P5 b9 d3 R2 k7 |
        , Y8 M! S- f5 F6 ^. t
    return this;
% t3 v: ~# B6 f2 w$ L% {  }
您需要登录后才可以回帖 登录 | 注册

本版积分规则

QQ|Archiver|手机版|SimulWay 道于仿真   

GMT+8, 2026-8-17 07:31 , Processed in 0.013604 second(s), 11 queries .

Powered by Discuz! X3.4 Licensed

© 2001-2017 Comsenz Inc.

快速回复 返回顶部 返回列表