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本帖最后由 zorsite 于 2015-10-9 20:49 编辑
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9# 慧娴亚伦
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总结的很有道理。" T* B8 r7 S+ Y6 s! }* `
也很感谢各位非常有意义的讨论和回帖。3 v' e2 c0 w0 }/ ^7 t% m
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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#pid17648
1 x9 {( g' |4 a) A3 F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:- " H, v9 Y4 h7 s2 m5 W
- int nrports = nrip(current);
# U3 t. \ _9 E5 G) S - int domax = 0;- C* Z1 X6 K) J/ w, V
- if (!inputopen(current)), }* h& C1 E4 {% V" l4 a% }
- return PULL_REEVALUATE_ON_READY;% _. w6 S! n6 t4 l: A2 g4 O
- double bestval = domax?-10000000000:10000000000;
' ^) p4 X7 m; ?/ D& Q1 Y5 p# b - int bestportnr = 0;8 W, r: \% q) j
- treenode bestitem = NULL;
, W a! T4 X7 r1 X! @) R! I - for (int i = 1; i <= nrports; i++) {
3 b% Z$ I8 r) S$ P - treenode object = inobject(current, i);
- Q5 [4 E* c; T, ] - if (!outputopen(object))
- ~; E9 ?% a! w! _$ U" ] - continue;, p& P F7 ^& Z; `. M* H
- int opnr = ipopno(current, i);2 r5 i; t/ r8 R
- if (!opopen(object, opnr)): Q. e P% [1 f4 ~
- continue;9 F- U- T* y+ j& v4 E5 R+ i: u6 |0 V
- int startindex = 1; int endindex = content(object) + 1; int incr = 1;
" U/ e, Y6 g+ @ - if (getvarnum(object, "lifo")) {7 U: i& E: g' ~3 d2 H- t. z
- startindex = content(object);
) S. {9 f6 n- |1 S - endindex = 0;! c9 T5 y! g' H& B; R* h
- incr = -1;8 R. E+ ^7 D. W7 k* V
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1 I4 U1 p* @# ` B; D8 @- for (int j = startindex; j != endindex; j+=incr) {4 M6 w- M4 f* T( M# W
- treenode item = rank(object, j);
! Z0 r( Z( W- c8 G! j1 e, i - if (evaluatepullcriteria(current, item, i)) {% \6 {& [4 G# C q' {
- double val = getitemtype(item);9 ]- }* K: M r* ?% B9 a. d* Q
- int newbest = domax ? val > bestval : val < bestval;/ O; E) D# i0 |& N+ Z* Q
- if (newbest) {4 D) L& {: Z$ i: k, h2 ~
- bestval = val;4 e5 n1 \% A6 ^
- bestitem = item;* A( N5 }" J$ n; g4 a3 C
- bestportnr = i;
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- }
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- pullitem(current, bestitem, bestportnr, BYPASS_ALL);
2 @; ~& I: l2 W- L5 M7 e - else return PULL_REEVALUATE_ON_READY;
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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。3 M; `4 i' `/ c6 p' s9 g
if (!inputopen(current))
$ I( S. g8 }* `7 K- \" C( G/ Q+ ereturn PULL_REEVALUATE_ON_READY;
) `5 H- n+ d t//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++) {
5 n1 y; p* D/ ?treenode object = inobject(current, i);
3 S I" n' L8 g7 cif (!outputopen(object))
4 m2 M) e+ x( `9 ?# E' A continue;
! z# n- c+ n# Z7 J//如果上游输出端口没打开,用continue跳出此次for循环+ A/ I2 D, ^3 f
int opnr = ipopno(current, i);8 | w1 ~/ u$ j5 a; d
if (!opopen(object, opnr))0 X3 l+ E+ m1 l% F
continue; M! G) {. t# o
//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& M$ m+ a7 V. O( i3 m, O) q
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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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