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Accessing bits in DWORD or DINT [SOLVED]

Created by: koaladua at: 12/11/2014 11:12 PM (11 Replies)

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12/11/2014 11:12 PM	Rate (0)
koaladua Posts: 10 Rating: (1)	Hello, Does anyone know how to access each individual bit in TIA PORTAL for a tag with DWORD or DINT as a data type? I want to use DWORD/DINT as an alarm group and set/reset individual bits accordingly. Thank you.
koaladua Posts: 10 Rating: (1)	Last edited by: koaladua at: 12/12/2014 9:04 PM
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12/12/2014 7:42 AM	Rate (3)
William_B Posts: 5225 Rating: (1192)	The short answer.. (slicing) #MyDword.x7 // bit #7 from a DWORD. ..but that is for the S7-1x00 PLC. ----- Also look at [LINK], the FAQ links are important for you. If the solution does not work for you - please be so kind as to refresh our memory: what PLC? w
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12/12/2014 9:03 PM	Rate (0)
koaladua Posts: 10 Rating: (1)	William B. The short answer.. (slicing) #MyDword.x7 // bit #7 from a DWORD. ..but that is for the S7-1x00 PLC. Spot on! I am using S7-1200. Sorry that I forgot to mention before since I didn't know that this feature only applies to S7-1x00. William B. Also look at [LINK], the FAQ links are important for you. Thank you for the link. Although I have to say I might need to spend more time reading about slicing and overlaying after this (such a beginner I am!!)
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12/12/2014 9:57 PM	Rate (0)
koaladua Posts: 10 Rating: (1)	William B. That means the SWAP function is required to have the two methods work the same: #value.%X{index} <-----------------> #b2[ #index ] Sorry if this question is too basic, but is there any reason why the swap is needed? Is it because of the structure of DWord? I once heard from a colleague of mine that the bits structure (generally in Siemens PLC) for DWord is not b3-b2-b1-b0 but it's somehow like b2-b3-b0-b1. Shed a bit of light on this please? Note: really llike the jpg and note! Thank you!
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12/13/2014 4:09 AM	Rate (1)
William_B Posts: 5225 Rating: (1192)	koaladua William B. The short answer.. (slicing) #MyDword.x7 // bit #7 from a DWORD. ..but that is for the S7-1x00 PLC. Spot on! I am using S7-1200. Sorry that I forgot to mention before since I didn't know that this feature only applies to S7-1x00. William B. Also look at [LINK], the FAQ links are important for you. Thank you for the link. Although I have to say I might need to spend more time reading about slicing and overlaying after this (such a beginner I am!!) Hi. Do not worry that the SLICING method is unknown to you. This is a subject that is not known by too many, as it is. The S7-1x00 has made some changes in the way the programs can be built and structured. I would suggest that you get the System Manual and Easy Guide for the S7-1200. Have a glance through the manuals and just get a 'feel' for what is written there. There is a chapter about SLICING too (System Manual). You can also consider getting the TIA Step7 Basic System Manual (that is a mouth full !). The trouble is I do not know a good guide from Siemens that builds up the knowledge of a fresh beginner. Probably that is why Siemens still have (real life) training courses. Cheers. w
William_B Posts: 5225 Rating: (1192)
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12/13/2014 4:27 AM	Rate (1)
William_B Posts: 5225 Rating: (1192)	koaladua William B. That means the SWAP function is required to have the two methods work the same: #value.%X{index} <-----------------> #b2[ #index ] Sorry if this question is too basic, but is there any reason why the swap is needed? Is it because of the structure of DWord? I once heard from a colleague of mine that the bits structure (generally in Siemens PLC) for DWord is not b3-b2-b1-b0 but it's somehow like b2-b3-b0-b1. Shed a bit of light on this please? Note: really llike the jpg and note! Thank you! Hi. When you have worked with other PLC types, this can cause much frustration. The link about slicing has an illustration that will help. The bytes are assembled from left to right, so MD14 will occupy bytes as follows.. MB14 - MB15 - MB16 - MB17 Each byte is constructed from 8 bits from right to left, example MB17.. M17.7 - M17.6 - ...... - M17.0 Now if we have MD14 = 1 it will means that M17.0 (least significant of the DWORD) is TRUE. Now MD14 is special in the sense that it is possible for us to deconstruct is into its smaller parts. Not all data allow this for us. Therefore we must use other methods. This methods can also be used for our example. Assume #value has been declared as symbol for MD14. #value.X0 is another method to extract a bit from the variable. In this case however the M14.0 bit is effected. That becoomes a problem for us when working mixed with MD14 and with its bits Mx.y and #value.X(?). Does this explaination help? w
William_B Posts: 5225 Rating: (1192)	Last edited by: Min_Moderator at: 12/15/2014 2:07 PM Delete link
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12/15/2014 9:44 PM	Rate (0)
koaladua Posts: 10 Rating: (1)	William B. The bytes are assembled from left to right, so MD14 will occupy bytes as follows.. MB14 - MB15 - MB16 - MB17 Each byte is constructed from 8 bits from right to left, example MB17.. M17.7 - M17.6 - ...... - M17.0 Now if we have MD14 = 1 it will means that M17.0 (least significant of the DWORD) is TRUE. w Going back to the example, how is it that #A1[24] is the one assigned value of 1? Why not A[31]? Thank you.
koaladua Posts: 10 Rating: (1)	Last edited by: koaladua at: 12/15/2014 9:49 PM
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12/16/2014 7:40 AM	Rate (0)
William_B Posts: 5225 Rating: (1192)	koaladua William B. The bytes are assembled from left to right, so MD14 will occupy bytes as follows.. MB14 - MB15 - MB16 - MB17 Each byte is constructed from 8 bits from right to left, example MB17.. M17.7 - M17.6 - ...... - M17.0 Now if we have MD14 = 1 it will means that M17.0 (least significant of the DWORD) is TRUE. w Going back to the example, how is it that #A1[24] is the one assigned value of 1? Why not A[31]? Thank you. Hi. The trouble comes in that the orientation (order) of the bits are right to left. That orientation is not disturbed. Bit 0 is on the right, Bit 7 is on the left - whichever byte you are looking at. When using the SWAP instruction (from the example), only the bytes are swapped around, the bit orientation remains in place. The crazy thing is at the start that Byte 0 is on the left in the DWORD and Bit 0 is on the right of that Byte 0. I know.. it is very confusing - but it is important to understand before attemping to deconstruct structure into individual bits. It is also important when working with alarming in HMIs. Please direct your attention to the FAQs I have spoken about earlier. Please play around with experiments on your PLC to get a feel for it. You are however more than welcome to return with questions afterwards. w
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12/16/2014 8:14 AM	Rate (3)
Kahl Platinum Member Joined: 12/16/2012 Last visit: 9/13/2024 Posts: 653 Rating: (130)	William B. ..... The crazy thing is at the start that Byte 0 is on the left in the DWORD and Bit 0 is on the right of that Byte 0. I know.. it is very confusing - but it is important to understand before attemping to deconstruct structure into individual bits. It is also important when working with alarming in HMIs. .... w This is related to the addressing method used with the hardware memory of the PLC (or any other system with a memory) There are two type of addressing methods: Big-Endian and Little-Endian. Here is a good explanation: http://en.wikipedia.org/wiki/Endianness I was able to memorize it by using the attached picture :-) Attachment big little endian.jpg (1676 Downloads)
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12/16/2014 8:49 AM	Rate (0)
Jens_app Gold Member Joined: 1/5/2011 Last visit: 6/15/2024 Posts: 815 Rating: (122)	Kahl William B. ..... The crazy thing is at the start that Byte 0 is on the left in the DWORD and Bit 0 is on the right of that Byte 0. I know.. it is very confusing - but it is important to understand before attemping to deconstruct structure into individual bits. It is also important when working with alarming in HMIs. .... w This is related to the addressing method used with the hardware memory of the PLC (or any other system with a memory) There are two type of addressing methods: Big-Endian and Little-Endian. Here is a good explanation: http://en.wikipedia.org/wiki/Endianness I was able to memorize it by using the attached picture :-) Thanks for that Kahl, it's always nice to know the history of something before just assuming it is without any background information. Very informative! Thumbs up So we can assume Siemensusesbig-endian system.
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