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5仿真币
在使用Order List时,当一个order list满足不了order的数量,要用第二个甚至第三个order list,有一段程序,请大虾解释一下,多谢!程序如下:
) A" Q* h& _4 D4 W; X* O# e3 Ebegin P_Order arriving procedure3 Y8 U$ ?$ e4 ?0 v b7 w
set V_NumWanted to 10& U; G9 x; e8 V, H$ B$ d5 `+ i
set V_NumOnList to OL_List1 current value# R+ E$ R! z6 z. c( e; {
order V_NumWanted loads from OL_List1 to P_NextProc
/ x: \7 c( w$ T3 E' p6 \ in case order not filled
/ d7 U. \8 [2 I* _8 k begin
; }7 @# ~+ h6 Z wait for .01 sec4 R* g( [/ t3 [! x$ b3 {
decrement V_NumWanted by V_NumOnList - OL_List1 current value# b3 s" \, E; @1 m* V- g
set V_NumOnList to OL_List2 current value' j. b3 [ D- r4 k
order V_NumWanted loads from OL_List2 to P_NextProc) V1 ?' d: ^; [' ]1 k4 {7 q p, Z
in case case order not filled
o# v o+ _, `- J4 A .................................................................# \6 Y1 S) _% [, L' z
" ]) T$ v1 e# `! T' H4 W
' W% b: V0 ~; L" \! N+ V. O0 Q我就是不明白上面用红绿色作标记的那句,为什么后面要加-OL_List1 current value.用前面绿色部分,即把不够的load算出来就不行了吗? |
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由于有in case order not filled这句,所以我觉得红色部分应该可以去掉。你还是自己测试一下吧,不一定要按书上说的做。
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