3/25/2020 8:26 AM | |
Posts: 4 Rating: (0) |
Hi, I need help to understand following code. As 'call' and 'Endcall' is not part of STL Syntex. Thanks. Call BLD 3 = L 24.0 CDB OPN DI 21 TAR2 LD 20 LAR2 P#DBX 0.0 UC FB 21 LAR2 LD 20 CDB BLD 4 End Call ------------------------------------------------------------------------------------------ |
Last edited by: amursaleen at: 03/25/2020 08:27:49Last edited by: Jen_Moderator at: 03/25/2020 11:06:23New subject after splitting |
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3/31/2020 2:15 PM | |
Joined: 12/9/2010 Last visit: 10/8/2022 Posts: 623 Rating: (77) |
Hi, BLD, - null instructions. CDB / swaps contents DB and DI registers.// Changed after reply. OPN DI 21 /Open DB21 in DI register. TAR2 LD 20 /Transfer AR2 to LD20 , local/temp memory dw 20 LAR2 P#DBX 0.0 /Move pointer for address DBX0.0 address to AR2 UC FB 21 /Open FB21, DB21 willl be referred as instances LAR2 LD 20 /Move Local/temp dw20 value to AR2. Hope it helps you ! If you could share entire logic, then only meaningful logic can be given.
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Last edited by: Sachinpatke at: 04/01/2020 13:33:25If you like the comment, rate it (right top) or thank it (left bottom). |
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4/1/2020 11:00 AM | |
Joined: 9/23/2005 Last visit: 10/18/2024 Posts: 4760 Rating: (728) |
CDB swaps contents DB and DI registers. In this context it is used to preserve current DI around the FB call. OT: Asking the same question over and over in new topics may not be the most efficient method to get the answer. Being you, I'd keep asking more specific questions in one topic till I had full understanding. |
Last edited by: jacek d at: 04/01/2020 11:07:49Regards, |
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