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本帖最后由 zorsite 于 2015-10-9 20:49 编辑 ' }) V0 J, f. V# N0 ?
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9# 慧娴亚伦
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, A3 d- a' f2 K6 W+ D4 N' K2 ]3 L总结的很有道理。
% W7 g4 } _5 f也很感谢各位非常有意义的讨论和回帖。- O& ^5 x% s2 N# ~0 J
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我之所以提出这样一个问题,是在做FMAT样题第一题的时候,FFFrenk版主提醒说有更简单的解决方案,就是系统自带的拉入策略“Pull Best Item”。missman也给出了自己的代码。见http://www.flexsim.asia/viewthread.php?tid=5361&page=1&fromuid=610#pid176485 a; h. _- R8 S! M! i/ T
于是我就去研究“Pull Best Item”策略,发现其设置了非常严格的端口检测条件,有兴趣的可以观摩观摩:- ) z @8 D7 ]% H+ h8 V+ ~5 t2 K$ f
- int nrports = nrip(current);
0 i; m* w" h- k0 k - int domax = 0;: V3 B# N4 A, p
- if (!inputopen(current))8 D) P& K x8 |! b; ^. R; L
- return PULL_REEVALUATE_ON_READY;
' b2 }3 Q/ H7 k, P* i - double bestval = domax?-10000000000:10000000000;
& _, |4 w& D! G1 ~) `+ u - int bestportnr = 0;
- F2 b8 u# V$ w/ u2 t - treenode bestitem = NULL;
; j' b5 h: G/ W8 A1 a0 ^* b - for (int i = 1; i <= nrports; i++) {* N1 S7 p& e$ K+ ?
- treenode object = inobject(current, i);
- a7 h- C. B% ` k( Q - if (!outputopen(object))
/ Y# X$ G3 n u( f( A+ {1 v6 J - continue;
+ N0 j0 r: A* i4 @+ h |& a/ f i - int opnr = ipopno(current, i);1 I6 g% m, s$ t
- if (!opopen(object, opnr))
2 Q. Z3 m1 }/ k# `2 V' M) e+ b - continue;
3 Y2 y0 l$ r- O - int startindex = 1; int endindex = content(object) + 1; int incr = 1;
; p7 [" b) B. M* X+ \) M& _# o - if (getvarnum(object, "lifo")) {2 r$ R2 ^9 ]$ e9 R! J
- startindex = content(object);
: s% w+ c- q: x: E K+ m - endindex = 0;
3 o. c9 K0 }* t$ [# e - incr = -1;1 D8 I, }5 t7 K- Z6 b8 Y
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4 L9 \, ^" N2 @& S8 `- for (int j = startindex; j != endindex; j+=incr) {
( M, `( L$ e8 ~8 J l( ` - treenode item = rank(object, j);
! l& _. {3 t( [* C4 D - if (evaluatepullcriteria(current, item, i)) {+ k4 V) u; [: d
- double val = getitemtype(item);5 M+ }' N% s* b% t( Z
- int newbest = domax ? val > bestval : val < bestval;
/ [! m8 @( j( Q2 \4 G - if (newbest) {6 Q( k! |& w9 u. } ?2 o
- bestval = val;
: F8 P" n+ w" e7 t: F - bestitem = item;+ v B! t% |$ H6 ?; {. L; G( G
- bestportnr = i;
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- if (bestitem)1 J4 Z8 V0 z0 g& u' c. H
- pullitem(current, bestitem, bestportnr, BYPASS_ALL); K/ M# Y0 V' W6 d' P
- else return PULL_REEVALUATE_ON_READY;
4 K% {, {5 a# y# |, I$ d - return 0;
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复制代码 在这段代码中,前三个if都是判断端口状态的。
& {; t$ V' W9 G9 vif (!inputopen(current))- u3 q0 [- t; s% c
return PULL_REEVALUATE_ON_READY;
) F; T S/ b6 x# @: z9 A//如果处理器端口没打开,用return跳出此段代码
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for (int i = 1; i <= nrports; i++) {
9 Q# G, e! m- ttreenode object = inobject(current, i);
% A* p- ~8 B' L0 S- Wif (!outputopen(object))
; E% T% ~+ a% o* R# q continue;
' W; c- W" C& w! S//如果上游输出端口没打开,用continue跳出此次for循环
! h( N: H4 D% s6 ~, Uint opnr = ipopno(current, i);
7 W: @# X4 @& J* o8 oif (!opopen(object, opnr))
: h: h# p% q; _' r$ h5 o$ @continue;4 Q" @1 J5 T# W$ K$ U
//如果上游输出端口打开了,继续判断上游实体与当前实体相连的那个输出端口是否打开,如果没打开,则用continue跳出此次for循环 B, L% N- `6 X6 V
. l' R' I8 C1 N; l: Y& {我主要是对最后一次判断比较不理解,因为根据我在前面的分析,固定实体是不能单独打开或关闭某个端口的。 |
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