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本帖最后由 zorsite 于 2015-10-9 20:49 编辑
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}; @( H. w$ G* @7 A$ v# b9# 慧娴亚伦 * e( s+ S$ B+ Y
7 `- w# k) N0 Z3 E( k! I总结的很有道理。* v2 q$ A4 H0 E! n7 q
也很感谢各位非常有意义的讨论和回帖。5 U. y+ v" n: i6 [. W0 |$ A1 z0 q
/ @. M6 ~3 h0 o9 Q. H* d5 D2 Z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#pid17648
7 ~( R; o2 ^4 G) \$ P/ R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:- 9 r7 O) `* `& d/ |: T4 w( b
- int nrports = nrip(current);
E. W6 H# G5 ~* o9 d - int domax = 0;
% L. G2 s d* t7 o3 f7 j - if (!inputopen(current))) k1 G# x9 `% O
- return PULL_REEVALUATE_ON_READY;
7 ]- O$ R5 T$ {6 p2 P5 o - double bestval = domax?-10000000000:10000000000;" U9 a* F9 r+ H( M+ |: I
- int bestportnr = 0;; A6 G, D, ^( H2 w
- treenode bestitem = NULL;" p, _, ~, m9 x6 V Y; k' r$ A
- for (int i = 1; i <= nrports; i++) {
/ [4 D" x1 p- p4 {4 ~( R - treenode object = inobject(current, i);
4 l+ c- i% ]& T( E5 T9 `5 ? - if (!outputopen(object))
* ?: p# x1 l0 [7 l) S - continue;3 Y, C6 ^8 k$ i7 B( O: Q2 R4 e, `9 {/ c
- int opnr = ipopno(current, i);' q4 t+ ^1 V3 J T7 Z4 e) b
- if (!opopen(object, opnr))7 z4 \3 y! f' z3 x. g6 n
- continue;! q$ J3 P' v; `+ \, H' b
- int startindex = 1; int endindex = content(object) + 1; int incr = 1; y! K0 Y3 w4 L3 t
- if (getvarnum(object, "lifo")) {5 l5 u9 k, V3 F5 B+ K- Y/ s% J4 m
- startindex = content(object);
: a$ X& L( B5 a0 K" P. } ?; O - endindex = 0;
9 d, S# N& I2 K6 w8 `3 A4 A k - incr = -1;
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- for (int j = startindex; j != endindex; j+=incr) {
1 Y. K Q8 z% s9 X0 M; p9 a - treenode item = rank(object, j);" X( @) o+ Q" c6 Z' q# N: Y5 b! x
- if (evaluatepullcriteria(current, item, i)) {5 y& U3 a9 Q, l/ i y6 v' }" `/ M
- double val = getitemtype(item);
/ y" r3 ^+ }+ x$ K% q* s2 b - int newbest = domax ? val > bestval : val < bestval;
, Z, z u2 O; U - if (newbest) {
5 W6 i7 z3 Y) h# Q* c9 v: G' G - bestval = val;/ l$ e. M% [0 ~5 [5 R
- bestitem = item;
% L2 X6 U) R* _+ X, @$ A" c9 H - bestportnr = i;
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- }* Z; `' n, h3 d$ U& o5 y4 b
- }7 u# X' E7 {: G
- if (bestitem)
7 X" c, u9 l8 G0 o/ y# A - pullitem(current, bestitem, bestportnr, BYPASS_ALL);. l; _* ]" m; m
- else return PULL_REEVALUATE_ON_READY; d; K) Z: l2 r8 z- O
- return 0;) [1 t" a5 u- i2 l: Q! d
复制代码 在这段代码中,前三个if都是判断端口状态的。# o. Y @2 F% h
if (!inputopen(current))) {" P9 s: ^. x! S- e: }" {
return PULL_REEVALUATE_ON_READY;
2 t$ }# _; { o; n& {6 h! F//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++) {
2 O$ Y6 I3 ^- w! W% ptreenode object = inobject(current, i);
" ~( ~0 e/ g, t% b0 |/ @if (!outputopen(object)), M% ~: v# R- p( @2 v: x6 K
continue;
5 J$ n: [0 n3 M, Q//如果上游输出端口没打开,用continue跳出此次for循环
$ N/ B! E4 `" i( B* Q; P6 y0 I- Kint opnr = ipopno(current, i);
' h X' W4 Q8 Z) s& Eif (!opopen(object, opnr)), ^ |4 g% S3 q: Y. n9 Q- z+ g
continue;- g) \" n) P& o5 K7 N) T
//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! N! X+ n, ?9 J- \4 f; d
* d5 j# |$ \7 y; h, [7 i4 p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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