现有一个S5程序，要转换成S7程序，可其中用到了DB0,这个如何处理？
下面的是相关的程序：
FUNCTioNFC16:VOID
AUTHOR:_991
NAME:SPEICHER
BEGIN
NETWORK
NETWORK
AM8.7;
AM247.1;
SM93.1;
JCM001;
JUM002;
M001:NOP0;
OPNDB221;
LW#16#0;
TDBW6;
TDBW8;
TDBW10;
M002:NOP0;
ANM8.7;
ANM247.1;
RM93.1;
NETWORK
OM94.2;
OM94.3;
OM94.4;
OM94.5;
OM93.1;
=M91.0;
NETWORK
OM91.0;
O(;
ANM169.0;
ANM169.1;
ANM169.2;
ANM169.3;
ANM169.4;
ANM169.5;
);
JCM003;
OPNDB221;
NOT;
SDBX10.6;
NOT;
SDBX10.6;
BEU;
M003:NOP0;
OPNDB221;
NOT;
RDBX10.6;
NOT;
RDBX10.6;
NETWORK
LMB8;
LW#16#82;
<>I;
JCM004;
//;
AM91.0;
JCM004;
//;
AM9.0;
JCM005;
OPNDB221;
NOT;
SDBX10.3;
NOT;
SDBX10.3;
JUM004;
M005:NOP0;
OPNDB221;
NOT;
RDBX10.3;
NOT;
RDBX10.3;
ANM91.0;
SM94.2;
M004:NOP0;
NETWORK
LMB8;
LW#16#83;
<>I;
JCM006;
//;
AM91.0;
JCM006;
//;
AM9.0;
JCM007;
OPNDB221;
NOT;
SDBX10.3;
NOT;
SDBX10.3;
JUM006;
M007:NOP0;
OPNDB221;
NOT;
RDBX10.3;
NOT;
RDBX10.3;
ANM91.0;
SM94.3;
M006:NOP0;
NETWORK
LMB8;
LW#16#84;
<>I;
JCM008;
//;
AM91.0;
JCM008;
//;
AM9.0;
JCM009;
OPNDB221;
NOT;
SDBX10.3;
NOT;
SDBX10.3;
JUM008;
M009:NOP0;
OPNDB221;
NOT;
RDBX10.3;
NOT;
RDBX10.3;
ANM91.0;
SM94.4;
M008:NOP0;
NETWORK
LMB8;
LW#16#85;
<>I;
JCM010;
//;
AM91.0;
JCM010;
//;
AM9.0;
JCM011;
OPNDB221;
NOT;
SDBX10.3;
NOT;
SDBX10.3;
JUM010;
M011:NOP0;
OPNDB221;
NOT;
RDBX10.3;
NOT;
RDBX10.3;
ANM91.0;
SM94.5;
M010:NOP0;
NETWORK
AM94.2;
JCNX000;
CALLFC17
(ANZ:=160
,LAEN:=90
,STRT:=M94.2
);
X000:NOP0;
//;
AM94.3;
JCNX001;
问题补充：
CALLFC18
(ANZ:=160
,LAEN:=90
,fbBA:=FC31
,FBSW:=FC27
,FBTW:=FC14
,FBSK:=FC29
,FBTK:=FC28
,STK:=M0.0
,FBST:=DB0
,BLH:=M0.0
,FBBH:=DB0
,LOB:=M0.0
,FBLO:=DB0
,EBRV:=M0.0
,FBEB:=DB0
,RES1:=M0.0
,FBR1:=DB0
,RES2:=M0.0
图片说明：
最佳答案
1,在你修改的程序中DB0块没有呀
2,在S5的程序中的DB0块是不被一般的块调用的,因为DB0是CPU系统的数据存储,不存用户信息数据.
提问者对于答案的评价：
只能读原程序了
