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本帖最后由 zorsite 于 2015-10-9 20:49 编辑 & Z* Q* C [: j
5 Q; C& P, Q, n1 Z9# 慧娴亚伦
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总结的很有道理。! x# K: w7 I0 i1 {2 [+ W- j0 o, T7 F
也很感谢各位非常有意义的讨论和回帖。
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, J) p7 k8 K: C- J: s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
6 B) O" M% M* a {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:-
( j( l; v- q( f/ Z6 K) w - int nrports = nrip(current);
: ~0 k6 ?( ]0 H' @, E; Z1 d - int domax = 0;
# t2 ]' s2 }( v2 L1 L - if (!inputopen(current))( l5 \" U# ?" y" q1 R
- return PULL_REEVALUATE_ON_READY;
9 P3 X9 a5 H; w n - double bestval = domax?-10000000000:10000000000;0 Z6 s: A9 v u, z+ {, H
- int bestportnr = 0;
j' \# C% d U. H( R, @. J/ X - treenode bestitem = NULL;
0 h+ A, w* z1 W - for (int i = 1; i <= nrports; i++) {
2 B- ]: L% j% @* W - treenode object = inobject(current, i);& ~" a! s3 R2 ]
- if (!outputopen(object))
5 }* n) F* Z# u; G - continue;
" d# E3 f1 M3 H& T: I, m7 N - int opnr = ipopno(current, i);
* G- H9 Q; V" Q0 f9 T" y - if (!opopen(object, opnr))( [( b# y2 l* z7 {5 ^
- continue;
' m# W3 b: i" X - int startindex = 1; int endindex = content(object) + 1; int incr = 1;
, G/ F/ v+ l. E" Z F% L- E, d - if (getvarnum(object, "lifo")) {! F# I% D1 M+ M0 |, g6 m
- startindex = content(object);
/ j/ I( Y& Q1 f, h - endindex = 0;
4 | m6 K& ~ T8 B" `5 w1 H - incr = -1;
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, |% I: i0 E1 O' \" G' `* K0 ~- for (int j = startindex; j != endindex; j+=incr) {7 R6 n- X% e! B, L5 c9 e
- treenode item = rank(object, j);
) d. ^& e, \; x. f - if (evaluatepullcriteria(current, item, i)) {
0 f* q) n% h5 T# n - double val = getitemtype(item);
8 L$ d$ x5 Y; o; Z5 K4 Z( W - int newbest = domax ? val > bestval : val < bestval; h: E; C& E% v! n) w
- if (newbest) {% L4 V9 a- E/ }6 U' Q( U0 b
- bestval = val;+ L$ s, |. k& G
- bestitem = item;
4 D3 U0 W7 Y0 g( _4 I4 Z4 P6 K$ f - bestportnr = i;8 ?" {2 N' {4 t; h" F) ?0 H3 Q
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- }
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- if (bestitem)
6 k& I% F) c; g* w1 C - pullitem(current, bestitem, bestportnr, BYPASS_ALL);3 ~9 q$ c* U; p+ u! R6 h7 `8 J6 X9 h
- else return PULL_REEVALUATE_ON_READY; _: v; l P- f' r# q' i, j
- 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。" A/ T S2 w( G5 Y4 R. Y# g4 |
if (!inputopen(current))# S4 R; U* U$ w' R- \
return PULL_REEVALUATE_ON_READY;
4 ^" y* _* w( b) U//如果处理器端口没打开,用return跳出此段代码6 T: Z- X* a& V9 N3 C
......
3 _ @4 C5 b. E0 ]& Hfor (int i = 1; i <= nrports; i++) { x+ w: [; A$ {" n' Y* ~
treenode object = inobject(current, i);3 m7 d- h: r4 c
if (!outputopen(object))
6 T# J4 ?* [5 Z/ `! K" U continue;
* ^% t# J* ~! ]+ B- I6 e) P" z//如果上游输出端口没打开,用continue跳出此次for循环
( {3 e9 n) S$ Y; qint opnr = ipopno(current, i);
# G8 D4 y- n5 V. H1 h- `1 x6 x2 [if (!opopen(object, opnr))
% u. _3 Z& C# o% J: [continue;
' K& G: @: k9 [% P//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
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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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