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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:
0 F! `7 x- v! O' n/ M" G$ ~( ]begin P_Order arriving procedure
+ X. @7 ^* ^6 u) j2 V set V_NumWanted to 10
/ J6 z! J: Y8 i Y; ^) o. a* r set V_NumOnList to OL_List1 current value% y; Q% x6 n9 x& G' n
order V_NumWanted loads from OL_List1 to P_NextProc
+ a, N4 f) }, N- e6 i% _, g in case order not filled
' T( c9 s' [6 c9 L6 u5 N' R begin
- N& a2 S& c6 q6 \/ s* S wait for .01 sec S9 I' s7 b( |' p2 L4 h& S
decrement V_NumWanted by V_NumOnList - OL_List1 current value
4 h/ J$ y2 g! Y( z2 c set V_NumOnList to OL_List2 current value3 I" O, f! V1 ~6 x' f) T
order V_NumWanted loads from OL_List2 to P_NextProc
, o* i! c( _" ^! d6 ]% u in case case order not filled
, @3 m: P0 H! M .................................................................
# V/ Q7 J7 {" @* ?3 x8 T3 s( k6 R
1 e7 z, Y1 B; n3 b3 l; x
& C" M5 L+ O I' f$ J) r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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