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S5 Block help

Created by: HTF at: 4/4/2019 9:15 AM (4 Replies)

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4/4/2019 9:15 AM	Rate (0)
HTF Regular Member Joined: 12/22/2014 Last visit: 9/11/2023 Posts: 3 Rating: (0)	Hello all I'm upgrading an old S5 to a S7-1200 and having trouble with some code understanding. In this segment, (see the Dos print attachment) on the connection between the 2 "AND" blocks, I have #F 51.1. I can't find any information on this "#" before the flag. There are quite a few places where this type of function is used. Can anyone please help? Regards, Helder Ferreira Portugal Attachment S5 Block.png (57 Downloads)

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4/4/2019 9:27 AM	Rate (0)
ixo65 Grandmaster Joined: 12/18/2014 Last visit: 10/3/2023 Posts: 30932 Rating: (3842)	The "#" is the sign for a so-called connector. It is used to store binary intermediate results. At S5, only one assignment can be made per network segment. If you now need a part of the network segment as, this connector is used. In your case, this makes no sense, since the AND block before only the F51.2 has as input. So F51.1 has the same state as F51.2 Attachment STEP5 V7.2 manual page 132.pdf (48 Downloads)
	Last edited by: ixo65 at: 4/4/2019 9:31:53 AM
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4/4/2019 9:52 AM	Rate (0)
HTF Regular Member Joined: 12/22/2014 Last visit: 9/11/2023 Posts: 3 Rating: (0)	That was just the simplest example in the program. Here I have #F 50.4 in segment 30. From what I understand then, there is no edge or other strange action other than making a 2 part function. In this case, F 50.4 can be read as "1" further on the program based on the OR logic and F 54.4 as "0" if F 52.4 is true in this case? This could be done normally in 2 segments? Attachment S5 Block2.png (59 Downloads)

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4/4/2019 10:13 AM	Rate (0)
ixo65 Grandmaster Joined: 12/18/2014 Last visit: 10/3/2023 Posts: 30932 Rating: (3842)	Yes, without the connector you must use two segments. Attachment Editor.pdf (51 Downloads)

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4/4/2019 10:54 AM	Rate (0)
HTF Regular Member Joined: 12/22/2014 Last visit: 9/11/2023 Posts: 3 Rating: (0)	Thank you very much. I was worried if there was some hidden, more complex function. Regards, Helder Ferreira

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