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本帖最后由 soso001 于 2014-12-17 14:53 编辑 : s& V) e# R6 @. e: c6 l5 J
; ?9 i9 |8 |6 A$ @不好意思,打错了个字母,呵呵呵!- c; F0 @$ ~6 S$ K r
大家好,本人em-plant新人,在看《生产系统仿真》第六章时,按书上的例子,写了段method,和书上是一样的,但总是提示
9 w8 S0 u# \: o1 ]“行19中错误,循环计数器必须为整数类型的局部变量”,但我的全局变量是 integer 类型啊,不知道怎么回事,求帮助!
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j,k,nstep,c, scheduleteps,mincomptime:integer;
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selecTab:table[integer,integer,integer,string];
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initorders;
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scheduledorders.delete;9 v: }6 A5 S, J
seletTab.create;2 G# J' Q) N2 c! K$ n- @+ P
4 {' v: G/ x/ V# ~ orders.initialize({2,1}..{2,*},0);
7 ~. w) d/ E7 s2 c4 V orders.initialize({4,1}..{4,*},0);
v9 v8 S* n' W" _& l" n9 u5 m for j:=1 to numorders loop
1 k; R, t, ]0 p* c, z' D orders[5,j]:=orders[1,j][2,1];3 ~% x+ b8 q- B3 y# p2 `; R
next;! S3 G2 G. r8 Y- y
occres.initialize({1,1}..{1,*},0);9 m: y! @) N A3 p& _
/ m9 @; [/ m; R4 S2 Q --XUNHUAN' L! x, ~) a$ {8 d: v1 ^
for scheduledsteps:=1 to numsteps loop --' \( n# l/ \. M8 e* A: C
selectab.delete;: ~0 Z9 x( `, {4 X5 b9 K( h
k:=1;
7 n! w; |) M( d) m/ i* B for j:=1 to numorders loop
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nstep:=orders[2,j]+1;" R! X$ w( A6 }( a; n- C
if nstep<=orders[3,j] then
' q. y% o3 \' \ selectab.writerow(1,k,j orders[4,j],orders[5,j],orders[1,j][1,nstep];
) @& O% h4 X( W$ X( n k:=k+1;
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& o5 j+ X- X# ~9 w2 _& K( k end;
: X, }; D$ T K6 R: v next;
& Z+ P" X, I& _- T( q/ j mincomptime:=selecttab.min({3,1}..{3,*});
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from j:=1 until j>selectab.ydim loop
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if selectab[2,j]<mincomptime then
7 @/ L1 X& L; P3 T j:=j+1;
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selecttab.cutrow(j);4 s2 h+ ~' Z' P
end;
5 O' \! g' v$ m% S- G& I9 a- Q end;
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j:=selectionrule(selecttab);
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nstep;+orders[2.j]+1;
2 ?5 H( ^& ?9 Y- a0 l orders[2,j]:=nstep;/ o, t, K1 \* b
m:=orders[1,j][1,nstep];
7 e( j# O3 O0 M! z% ]+ k c:=orders[5,j]
5 S9 k+ l4 j( f+ f$ S" Z occres[1,m]:=c;
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) N* N, ^7 z3 Y1 _, B --jieguo:start,end,res,job,step
$ G6 a, R; b$ z& s3 E scheduledorders.writerow(1,scheduledorders,orders[4,j],c,m,j,nstep);8 i: [4 p& \( Z, X6 S0 U
for k:=1 to numorders loop! x1 V$ p- n% W, C) l
nstep:=orders[2,k]+1;
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/ J3 y) o, H; y( i+ Z4 S$ A if nstep<=orders[3,k] and orders[1,k][1,nstep]=m
" o3 m' x7 R7 J6 s0 v5 l, E --
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orders[4,k]:=max(c,orders[4,k]);2 o) [2 C- `: j8 j( F; v' x7 l$ F
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orders[5,k]:=orders[4,k]+orders[1,k][s,nstep];
}- @7 w; ^% b% U1 o0 U --
2 C/ i& w# O# I! c' t0 K K5 d end;) E% I1 x; u: ]* x
next;) _' v" C& ~" V/ |
nstep:=orders[2,j]+1:
1 `. C0 t! w! T4 O& K4 L if nstep<=orders[3,j]0 x+ ?) C4 P7 }3 u) g1 D
4 r" ]; l# P, D x) V then
) `2 P/ I! z# n* x; v orders[4,j]:=max(c,occres[1,orders[1,j][1,nstep]]);: o; r9 X' A7 ~% e N, B5 _
orders[5,j]:=orders[4,j]+orders[1,j][2,nstep];! f4 [) q @( _4 K6 V' O
end; t9 Q- M! i& n$ i" i5 g. D
next;
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; N8 D( q" d& K; X print"ready.duration of the schedule:",scheduledorders.max({2,1}..{2,*});8 ?2 X1 q2 I3 i \# V, p
mygantt;- ^) J" k2 r4 m& k
ens;
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