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本帖最后由 soso001 于 2014-12-17 14:53 编辑
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不好意思,打错了个字母,呵呵呵!7 N X' O+ O2 a+ }' n' Z
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- n& ^, ], E, x4 g c6 s- B
“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!, u* K! h& V: T, R
is
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M:string;0 e) H+ L5 X: ]1 E
selecTab:table[integer,integer,integer,string];
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initorders;
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# a! g ]8 l, ~( T scheduledorders.delete;0 o! ]& W: ^1 {% K: `
seletTab.create;2 q b& v1 _( V- ?# I1 m
1 A4 T; f. Q' x3 ? orders.initialize({2,1}..{2,*},0);* M7 A! e, k) Y1 o. W4 w
orders.initialize({4,1}..{4,*},0);0 E6 E U6 E- f5 C# N+ n0 D3 }6 b) C
for j:=1 to numorders loop" ^- @0 `0 P( w) z; W
orders[5,j]:=orders[1,j][2,1];0 ~4 d7 a' d* j! c S5 n% r, P. U
next;0 `5 F7 x9 h" C
occres.initialize({1,1}..{1,*},0);
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--XUNHUAN
5 z6 { y7 U. |) b* f, ]1 ~ for scheduledsteps:=1 to numsteps loop --
$ \+ @* L) E- }$ G selectab.delete;
) B1 W, V! y3 |1 ]5 Q k:=1; m2 t9 U7 G/ Z$ z! m) |! K$ f1 A& M
for j:=1 to numorders loop
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nstep:=orders[2,j]+1;
3 A7 Z% a! _3 o# i if nstep<=orders[3,j] then
i/ C, z" T: E' K O' U selectab.writerow(1,k,j orders[4,j],orders[5,j],orders[1,j][1,nstep];0 f- `. U) [" b1 \7 a
k:=k+1;
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next;
& {+ Q+ v8 i8 B! {( C5 | mincomptime:=selecttab.min({3,1}..{3,*});$ h j p5 P9 x1 k- j) N* x
4 n) x( @ Z1 K9 ~5 T2 l from j:=1 until j>selectab.ydim loop
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: Z# M9 J( ?* U' Z9 z if selectab[2,j]<mincomptime then
8 K5 ~7 U: Q9 ^ j:=j+1;
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selecttab.cutrow(j);" e$ [" }! Y4 m1 ? o0 C8 l
end;5 L0 y. z' j9 }3 C6 `
end;
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j:=selectionrule(selecttab);3 \4 ~% |( e& I; g
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nstep;+orders[2.j]+1;
8 T; z: O' p, p6 J orders[2,j]:=nstep;3 \. Z7 F/ B$ n+ j
m:=orders[1,j][1,nstep];2 Q6 `6 \+ |& u9 v9 f" q
c:=orders[5,j]- p# b6 B9 a/ b& U2 h
occres[1,m]:=c;, [ S; r( R! z- D; j
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--jieguo:start,end,res,job,step
& Z& W) R0 y' G scheduledorders.writerow(1,scheduledorders,orders[4,j],c,m,j,nstep);
* S! r! ^2 V8 v E9 I/ y( D for k:=1 to numorders loop& ]& U% Z( M) `9 }+ h* ]
nstep:=orders[2,k]+1;
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if nstep<=orders[3,k] and orders[1,k][1,nstep]=m& G- L8 g+ x. ~" w d0 Y; \
--
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orders[4,k]:=max(c,orders[4,k]);1 w8 O: v1 `, z; }4 H
--
% s" K/ V, g( p( \) \8 c7 t \ orders[5,k]:=orders[4,k]+orders[1,k][s,nstep];% U0 g3 k9 K9 s, i# ^
--$ w5 t9 q% p9 X1 p0 Z7 i& _; Q
end;3 s- z6 R8 Y6 B) Y5 @0 T {
next;
7 ]3 A6 v% @- L! O+ n nstep:=orders[2,j]+1:, t2 l/ U5 ` p j, S) ]$ ]. v. X
if nstep<=orders[3,j]; L" @, f7 m% g! M/ G+ q+ B
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then
8 B# K5 i3 k3 U5 X' M0 ^% K4 x orders[4,j]:=max(c,occres[1,orders[1,j][1,nstep]]);
7 J& t% B& a# X% N, r( d* e orders[5,j]:=orders[4,j]+orders[1,j][2,nstep];
, ~8 _: T% D. i0 p% n end;
! c7 [% u: J' Q- L next;
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print"ready.duration of the schedule:",scheduledorders.max({2,1}..{2,*});
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ens;3 P8 P: _) Z, s$ ^0 W1 `0 b7 ~
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