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Logic operation in step7

Created by: Oscar_83PL at: 5/31/2011 4:45 PM (4 Replies)

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5/31/2011 4:45 PM	Rate (0)
Oscar_83PL Silver Member Joined: 10/22/2008 Last visit: 12/23/2025 Posts: 378 Rating: (5)	Hello, I am dealing with some logic operations in step 7. I am not sure if I am doing it right or there is some simpler ways. In FB I want to convert Array of bytes to bool. I do it in this way, but if I change in stats table there everything is messing up. [code] L #ArrayRecived[1] T LB 30 L #ArrayRecived[2] T LB 31 L #ArrayRecived[3] T LB 32 L #ArrayRecived[4] T LB 33 A L 30.0 = #Clip_Grey A L 30.1 = #Clip_Black A L 30.2 = #Footrest_RHD_M A L 30.3 = #Footrest_RHD_Oys A L 30.4 = #Footrest_RHD_Ant[/code] Is there a function that gives me back info that in byte there is at least one bit high and output it to me in bool format? Next thing is I want to set/reset one bit in byte tag. I wanted to use AND and OR blocks but it does not support byte format. Tell me how you cope with problem like that.

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5/31/2011 4:50 PM	Rate (0)
IBN-Service Posts: 8946 Rating: (1000)	A good idea will be to check OSCAT library. Download in english language possible! It is free of charge and will give you a lot of funtions and logics that you need.
IBN-Service Posts: 8946 Rating: (1000)	Last edited by: IBN-Service at: 5/31/2011 4:53 PM Link added
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This contribution was helpful to 1 thankful Users Oscar_83PL

5/31/2011 5:29 PM	Rate (0)
Niranjan Prasad Posts: 691 Rating: (37)	Since you are using STL Languange,you could have used some tricky way to tackle those problems Oscar_83PL Hello, Is there a function that gives me back info that in byte there is at least one bit high and output it to me in bool format? A( L #ArrayRecived[1] L 0 >I ) = L30.0 A( L #ArrayRecived[2] L 0 >I ) = L30.1 Oscar_83PL Next thing is I want to set/reset one bit in byte tag. I wanted to use AND and OR blocks but it does not support byte format. If you know the Address of the Byte , then Directly you can access the Bit like DIX Byte.Bit ( DIX 10.0)
Niranjan Prasad Posts: 691 Rating: (37)
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5/31/2011 5:52 PM	Rate (0)
Aret Posts: 2348 Rating: (265)	1) If you need at least 1 bit = 1 - see Shar's example. If you need certain bit (2nd in my example): L #Input_byte L B#16#04 // it's a mask - 0th bit=1, 1st bit=2, 2nd bit=4 etc. AW // yes, you perform AND WORDs, but that doesn't matter - it ANDs bytes also L 0 ==I JC zero // 2nd bit in Input_byte =0 JU one // 2nd bit in Input_byte =1 2) Changing bit is similar L #Input_byte L B#16#04 OW T #Input_byte 3) If you bit number is calculated inside programm you need SLW instruction like L 1 L #Bit_number SLW T #mask And then use #mask instead B#16#04
Aret Posts: 2348 Rating: (265)	Last edited by: Aret at: 5/31/2011 5:53 PM
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6/1/2011 8:19 AM	Rate (0)
Oscar_83PL Silver Member Joined: 10/22/2008 Last visit: 12/23/2025 Posts: 378 Rating: (5)	I do all the magic with masking bytes but while direct adressing with local variable it changes sometiomes its adress when you add new value to the table. And thats the main disadvantage of my try. Other way i have to convert all the time my tags from Byte to Word and in other drirection then and use symbolic adresses. In oscat lib. is everything I need so far.
	Last edited by: Oscar_83PL at: 6/2/2011 9:12 AM
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