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Need Support for Convertion S5 - S7

Created by: herumang at: 7/12/2010 6:40 AM (9 Replies)

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7/12/2010 6:40 AM	Rate (0)
herumang Posts: 6 Rating: (0)	Dear All, I have this code in S5 (135U) 928B : FB 180 D:030807ST.S5D LEN=248 Segment 1 Name :AN-TULO Decl :DB I/Q/D/B/T/C: B Decl :LUKM I/Q/D/B/T/C: D KM/KH/KY/KS/KF/KT/KC/KG: KF 000B :DO =DB 000C :LW =DB 000D :L KB 255 000E :AW 000F :L RS 34 0010 :+F 0011 :LIR 1 0012 :T DW 0 0013 :ADD BN +29 0014 :T DW 2 0015 :L DW 1 0016 :L KB 49 0017 :>=F 0018 :JC =M001 0019 :T DW 1 001A M001 :L DW 0 001B :L DW 1 001C :+F 001D :L DW 2 001E :TNW 20 =================================================================== This function Called in other FB with this code : :JU FB180 Name : AN-TULO DB :DB180 and this is some of DB180 default value 0: KH = 32F4; 1: KH = 0199; 2: KH = 3311; 3: KH = 0000; 4: KH = 064E; 5: KH = 8400; 6: KH = 0031; 7: KH = 0100; 8: KH = 0764; 9: KH = 0000; 10: KH = 0282; 11: KH = 0000; 12: KH = 2814; 13: KH = 1E42; ============================================================= Can anyone help to explain this code and how this code converted to S7 ? Best Regards, Herumang
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7/12/2010 10:03 PM	Rate (0)
jklm Platinum Member Joined: 12/22/2006 Last visit: 1/25/2022 Posts: 1742 Rating: (308)	Hello, The end of the program uses TNW. 001A M001 :L DW 0 001B :L DW 1 001C :+F 001D :L DW 2 001E :TNW 20 TNW20 transfers 20 words between two memory areas pointed by absolute addresses. TNW copies backwards (the transfer is executed starting with the highest address of the source area and is completed with the lowest address). The addresses used to point the areas are always the upper (higher) addresses of source and destination area. The end address of the source area must be loaded in accu 2 (sum of DW0 + DW1 in your program) and the end address of the destination area in accu 1 (content of DW2 in your program). These source and destination addresses must be computed, and that is what the rest of the program does. The following part computes the absolute memory address of the start of the DB and stores the result of DW0 of that DB 000B :DO =DB 000C :LW =DB 000D :L KB 255 000E :AW 000F :L RS 34 0010 :+F 0011 :LIR 1 0012 :T DW 0 after the following part, DW2 contains the absolute address of DW 29 Will be used as end address of the destination 0013 :ADD BN +29 0014 :T DW 2 This part checks if DW1 has a value larger or equal to 49 If less, 49 is stored in DW1 0015 :L DW 1 0016 :L KB 49 0017 :>=F 0018 :JC =M001 0019 :T DW 1 The source address (DW0 + DW1) points to the address of DW49 So, I believe that the program copies dw30 ==>dw49 into dw10 ==> dw29 from the DB specified as parameter DBNR. In Step7 you can call a simple block move SFC20 that copies the contents of a memory area (= source area) to another memory area (= destination area).
	Last edited by: jklm at: 7/12/2010 10:06 PM Best regards jklm Suggestion was helpful? Feel free to rate it.
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7/13/2010 7:10 AM	Rate (0)
herumang Posts: 6 Rating: (0)	Hi jklm, Thanks for your reply, but i still curious with the beginning code : 000B :DO =DB 000C :LW =DB if the absolute "=DB" is DB180, all transfer DW is inside DB180 ? and what is "LW =DB" does working ? and what is the actual value ? Thank you and regards, herumang
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7/13/2010 10:31 AM	Rate (0)
fritz Diamond Expert Joined: 10/7/2005 Last visit: 10/5/2024 Posts: 3024 Rating: (1054)	herumang .. but i still curious with the beginning code : 000B :DO =DB 000C :LW =DB if the absolute "=DB" is DB180, all transfer DW is inside DB180 ? and what is "LW =DB" does working ? and what is the actual value ? Hello herumang "DO =DB" opens the DB hat is attached to the formal operand "DB" of the FB, which isDB180 in your code code example. "LW" is command that loads the bitpattern of the formal operandinto accu1 Low (the lowest 16 bit of accu1). "LW =DB" is in your case used to get the the DB number "180". Sincethe bitpattern value for the "DB" part of "DB 180" is of no use, it is masked out with the next two STL lines ("L KB 255" and"AW"). The result is that accu1 has an Integer value of 180 in it. Just in case you don't have it yet, all commands and more are explained in the CPU 928 programming manual which you can download from HERE. P.S: Very nice explanation by jklmandI agree with his advise on what the logic does and only haveone little addition to it: DW 1 in has a value of 199hex (=409 decimal) in your example DB180. As 409 is bigger than 49 (the minimum allowed value that your FB checks for) this means that DW390->409 arecurrently copied to DW 10->29. I suggest you check if DW 1is changed elsewhere in the code as its value determines what source address range is copied into thefixed (DW 10->29) destination address range and ideally you test run the S5 logic in a 928 CPU to verify its functionality fully. I hope this helps and good luck with your conversion
	Cheers fritz The advice and opinion given in this thread is that of the author and does not necessarily reflect the views of I S Systems Pty Limited. To find out how I S Systems can help you with your automation needs please visit www.issystems.com.au.
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7/13/2010 11:35 AM	Rate (0)
herumang Posts: 6 Rating: (0)	Dear fritz, Well it opened my point of view, i will try to think harder to solve the rest of the code. Thank's all.. I'll contact you if i get stuck. Best regards, herumang
herumang Posts: 6 Rating: (0)
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7/15/2010 6:42 AM	Rate (0)
herumang Posts: 6 Rating: (0)	Dear All, i got stuck again with this code : 00C5 :L RS 34 *** 00C6 :ADD KF +255 00C8 :LIR 1 00C9 :L DW 11 --> default value is 0 00CA :+F 00CB :T FW 252 00CC :LIR 1 00CD :T FW 254 *** 00CE :A F 250.0 00CF := F 255.0 00D0 :A F 250.1 00D1 := F 255.1 00D2 :A D 7.9 -20:7.9 00D4 := F 255.3 00D5 :L FW 254 00D6 :L FW 252 00D7 :TIR 3 --> this one is i know, transfer accu-2 to accu-1 i got stuck with code betwen asterix (*). My perseption is like this (please correct if i'm wrong) : 00C5 :L RS 34 ---> start of DB0 00C6 :ADD KF +255 ---> add by 255 00C8 :LIR 1 --> it is load of DB 255 DW 0 ..?? 00C9 :L DW 11 --> default value is 0 00CA :+F ---> result of accu-1 is 255 because the value of DW 11 is 0 (zero) 00CB :T FW 252 ---> transfer value 255 to FW 252 00CC :LIR 1 ---> Load again ? 255 to accu-1 ? 00CD :T FW 254 ----> transfer to FW 254. so, is my perseption true..?? Thank you for your attention. Regards,
herumang Posts: 6 Rating: (0)
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7/15/2010 12:31 PM	Rate (0)
fritz Diamond Expert Joined: 10/7/2005 Last visit: 10/5/2024 Posts: 3024 Rating: (1054)	Hello again herumang nice effort and good on you for giving it your best shot, but I would have a slightly different take on on what yourcode snippet does. [code]L RS 34 // Load Start address of DB address list into accu1 ADD KF +255 // Add 255 to accu1 (= point to memory cell that contains the start address of DB255) LIR 1 // Load Register 1 (=accu1) with the contents of the memory cell whose address is in accu1. Result: accu1 contains thestart address of DB255 (to be precise, the absolute memory address of DB255's DW0) L DW 11 // Loada value tocreate a DW offset, we assume for this example DW11 in DBxyz has a value of 10 +F//accu1now containsthe absolute memory address of DB255's DW 10 T FW 252 // store thisaddress in FW252for later use LIR 1 // load the contents of DB255's DW 10 T FW 254 // store this value in FW254 A F 250.0 // change/manipulate some bits in FW254 = F 255.0 // . A F 250.1 // . = F 255.1 // . A D 7.9 // . = F 255.3 // . L FW 254 // load changed FW254 L FW 252 // load FW252 into accu1 (=the absolute memory address of DB255's DW 10) TIR 3 //Transfer the contents of register 3 (=accu2 Low=value of changed FW254) to the memory cell whose address is in accu1 (which is DB255's DW 10 in this example). [/code] To surmise, your S5 logic does the following: A Dataword from DB255 is loaded, some bits in it are changed and the changed value is written back into the same Dataword. Which Dataword is loaded depends on the value of DW 11. Note that your code snippet does not reveal in which DB this DW 11 and that the employed absolute address access does NOT open DB255 via the DB Registers. As you can see from this example, dealing with absolute address access in S5 can be pretty hair-raising stuff. While this addressing mechanism doesn't exist anymore in S7 (and there is really no need for it in S7 anymore either) I still strongly recommend that you get hold of a 928B CPU and test some of the blocks you are dealing (it greatly helps in verifying that the manully decoded functionality is understood correctly before you create your S7 equivalent for it). Have another look too at Chapter 8 and 9 of the previously supplied manual,they do contain a few examples concerning this and are worthwhile having as a printout while dealing with this kind of S5logic (I fear you will discover more of this kind of hardcore S5 logic). I hope this helps
	Last edited by: fritz at: 7/16/2010 4:13 AM Added summary Cheers fritz The advice and opinion given in this thread is that of the author and does not necessarily reflect the views of I S Systems Pty Limited. To find out how I S Systems can help you with your automation needs please visit www.issystems.com.au.
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7/16/2010 9:10 AM	Rate (0)
herumang Posts: 6 Rating: (0)	Dear Fritz, Thank you again for great explanation. I'm just wonder for this code, if it is a trick to access DB180 greater than 255 DW ? Because in DB180, there are 567 DW, so after converted in S7, DB180 has 1134 DBW. And there are no DB255 of DB structure in my program. Thank you & Regards
herumang Posts: 6 Rating: (0)	Last edited by: herumang at: 7/16/2010 9:19 AM
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7/16/2010 10:30 AM	Rate (0)
fritz Diamond Expert Joined: 10/7/2005 Last visit: 10/5/2024 Posts: 3024 Rating: (1054)	herumang I'm just wonder for this code, if it is a trick to access DB180 greater than 255 DW ? I wouldn't call it a trick dear herumang, but rather one way out of a few thatallowedyou to get access to DB Words past the magic 255 boundary that existed in S5 (DB's could always be longer than 256 Words, but 255 was the highest value you could access directly in your logic as only one Byte was available to store this address value). herumang And there are no DB255 of DB structure in my program. There better be a DB255 as it is what your logic is aiming for. Then again, perhaps the FBis not called orDB255 is created via logic on startup and does not exist in your program backup or I got my logic explanation wrong ...
	Cheers fritz The advice and opinion given in this thread is that of the author and does not necessarily reflect the views of I S Systems Pty Limited. To find out how I S Systems can help you with your automation needs please visit www.issystems.com.au.
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7/20/2010 9:52 AM	Rate (0)
herumang Posts: 6 Rating: (0)	Dear fritz, Thanks for your explanation. regards, herumang
herumang Posts: 6 Rating: (0)
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