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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。) g9 I' L/ ?3 V$ \0 h
也很感谢各位非常有意义的讨论和回帖。1 b2 f; L1 q" l
9 z8 X) o Y+ U& v! D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! N A6 z( m) V0 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:- ! \5 f0 c$ p9 D+ P
- int nrports = nrip(current);
- E! |' S3 x, R- F' i - int domax = 0;/ H+ D/ i' v5 y3 h4 c% L+ b
- if (!inputopen(current))
& T9 t7 j0 ~" r0 c1 O - return PULL_REEVALUATE_ON_READY;8 m4 i- P& m3 j$ ^0 J* U
- double bestval = domax?-10000000000:10000000000;: V- ^4 P, H/ z9 P
- int bestportnr = 0;. P! r$ p% ?( f' m
- treenode bestitem = NULL;: Y0 |8 I8 A! ^ H, f
- for (int i = 1; i <= nrports; i++) {
8 o( v2 h4 `2 O; t4 ~ a) A; c6 y - treenode object = inobject(current, i);. d! h9 t$ Q, T8 s
- if (!outputopen(object))+ g5 c0 E& A6 V$ a& M, _ d
- continue;
" f/ D- l6 Q- X$ t, s4 A - int opnr = ipopno(current, i);! I* K$ z1 b. ^6 c
- if (!opopen(object, opnr))9 z; E) G' u' N
- continue;* ?8 B+ A1 t) Y$ x! ~! o/ t' A
- int startindex = 1; int endindex = content(object) + 1; int incr = 1;
- v4 h8 ^; l+ Y0 k7 x0 w - if (getvarnum(object, "lifo")) {
( A5 T% d/ z9 o4 K - startindex = content(object);: {4 q) G" [. Y) O% Y3 H. q, x
- endindex = 0;& o: i) c, O# L4 }
- incr = -1;) T' r4 X; Y$ Z; E: F w f+ Z
- }
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6 X% K8 l* {" _. Q- for (int j = startindex; j != endindex; j+=incr) {
- H7 I" O8 I, Y9 v - treenode item = rank(object, j);2 k8 q4 ^* J2 O' n
- if (evaluatepullcriteria(current, item, i)) {0 r! j7 }+ R- S2 o# o: U
- double val = getitemtype(item);
7 s+ a9 _. f! Q( v; ? - int newbest = domax ? val > bestval : val < bestval;; _' P( B& A O4 s+ v
- if (newbest) {
: l/ d/ L% }2 L0 g - bestval = val;) S) R s& U" u; |% x
- bestitem = item;6 F! T- J5 b0 f; h* F) [0 L0 Z
- bestportnr = i;
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- }
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- if (bestitem)
# u5 M& |8 H% m: f0 P: V1 X - pullitem(current, bestitem, bestportnr, BYPASS_ALL);* `) e9 E1 D5 t2 b# f8 }
- else return PULL_REEVALUATE_ON_READY;
8 p2 ?' H7 M/ w5 I9 h2 l+ 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。
1 Z9 j4 L. }, ~if (!inputopen(current))
6 R" O9 q- }: r- _$ Ureturn PULL_REEVALUATE_ON_READY;
1 y% [/ c' a0 ~: y//如果处理器端口没打开,用return跳出此段代码
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3 C* q7 s+ c: { zfor (int i = 1; i <= nrports; i++) {, B- |3 C- n& C' M# h7 M
treenode object = inobject(current, i);, o' j3 d3 |# S* V
if (!outputopen(object))
8 `# A4 y S7 e: o. f* g" [1 R continue;8 b8 \, j! O! D( P# U/ P" @
//如果上游输出端口没打开,用continue跳出此次for循环
! |1 v- }. Z {int opnr = ipopno(current, i);
8 M4 X0 M4 ?! I( ^: \- _if (!opopen(object, opnr))9 f+ |) q( I. Q
continue;
$ p9 V+ S" m9 X/ a//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( Z' `2 k! f8 i+ w4 U( T
9 e" c6 i8 @9 ~4 X5 u% b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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