I am trying to write a .dbc file for a can-open data log (example of one of lines I am trying to use below)
Time 884.163000, ID:2a1, Data Bytes (0)7b (1)00 (2)95 (3)68 (4)e5 (5)8e (6)49 (7)54
I have written .dbc files using both motorola and intel endianness using candb++ covering 16 bit data over 2 bytes, but this has always been with sequential bytes, ie- (2),(3) or (5),(6).
The bytes I need to use for the particular data in the above example are (3) and (7) in intel format (54,68 in this case). I have written a .dbc for just byte 3 shown in the snip below.
BO_ 673 Rig_Pressure: 4 Vector__XXX
SG_ Pressure_Multiplex M : 15|8#0+ (1,0) [0|0] "" Vector__XXX
SG_ Pressure m0 : 31|8#0- (0.45,58) [0.399999999999999|115.15] "Bar" Vector__XXX
I am asking if there is a way to modify the text file (or use cabdb++) to specify each bit or pick 2 non sequential bytes in the .dbc, something like modifying the bit start something like
31|8#0- to 31|8#0- 63|8#0-
I am far from a computer programmer, I much prefer GUI based programs and am only starting out in learning python, so please be gentle!!!
Thank you!
Related
I'm reading in a stream of data, 64 bytes to be exact. I want to read 16 bits starting at the 480th bit of the incoming data. Unfortunately, I do not know what the incoming data type is, it's a bunch of random characters/boxes. Reading it in as an unsigned short (v), I get the number I am looking for, which for this example is 13.
my $satt_id = unpack("x60v1"), $msgdata); #$satt_id == 13
This results in $satt_id == 13, which is 00000000 00001101.
If I pull the data as 16 bits (b or B), the string does not reflect the value of 13, but rather is byte-swapped or reversed.
my $satt_idb = unpack("x60b16", $msgdata); #satt_idb == "10110000 00000000"
my $satt_idB = unpack("x60B16", $msgdata); #satt_idB == "00001101 00000000"
Why is this occurring? I want to alter the data and resend out the message, which would be relatively easy if all of the message elements were the same size (16 bits, just pack back as it was unpacked), but some are 6, 4, 2, and 1 bits. Should I just use little-endian b and then reverse? After altering the data reverse it back to original order and then pack it back as b?
Completely separate and not perl related, but this haunted me in a different utility. I just conceded by swapping the values in the Enum designation. It worked, just wasn't very viable when the amount of bits got higher than 4 (16 different values).
Thanks!
EDIT: I'm guessing this is just related to binary notation? Apparently starts from the right? So $satt_idb is correct, if you read right to left. So to make it more user friendly, just reverse, alter, then reverse again and repack?
EDIT2: Basically I'm trying to make a user-friendly method of editing messages coming through a data stream. As I mentioned in the comments, if I want to edit a single bit from 0 to 1 (which in the message represents something as true/false), I don't want the user to have to worry about editing the octet of data received, just select from a dropdown of true/false.
If it works with v, it means the data is in little-endian byte order, which means
0b0000000000001101
is stored as
0b00001101 0b00000000
which is what you got.
Should I just use little-endian b and then reverse?
No. You are likely doing something incorrect if you are converting the numbers to a text representation (binary).
If you did somehow want the binary representation of the number, you could use
sprintf("%16b", $num)
I'm using AT24C512 EEPROM which is 512KB along with my STM32
I'm able to write 128bytes of data at once using
HAL_I2C_Mem_Write(&_EEPROM24XX_I2C,0xa0,Address,I2C_MEMADD_SIZE_16BIT,(uint8_t*)data,size_of_data,100)
but the issue is that i want to write more data after the data that was just wrote, but the EEPROM will replace the data as the Address is the same
so how can i skip the written address ?
This answer is not about using HAL with I2C, but hope it will point you
Just check datasheet (I looking into STM32F0) and you can see that the limit is 255 bytes (register CR2:NBYTES), I'm not sure if there is another limitation in HAL, but using direct access to registers you can sent 255 bytes at once or fragment it and sent how much you want.
For fragmenting there is bit CR2:RELOAD, if you set this, then at the end will be not transfer stopped, and you can update next NBYTES, .. when you will set last block of bytes (which will fit into NBYTES) then clear bit CR2:RELOAD.
This has one disadvantage, that every 255 bytes, you will be interrupted.
i think you should check the AT24C512 datasheet page 7.
If more
than 128 data words are transmitted to the EEPROM, the
data word address will
“
roll over
”
and previous data will be
overwritten. The address
“
roll over
”
during write is from the
last byte of the current page to the first byte of the same
page.
I am trying to program an ISL12022M RTC and am having trouble interpreting the register map (self taught with little experience). The documentation says that the RTC registers (SC,MN,HR,DT,MO,YR,DW) are BCD representations. In order to allow write capabilitiy into the RTC registers the WRTC bit(bit 6 of address 08h is set to '1'.The map looks like this:
The FAQ example from the Intersil site tells me that to set the WRTC bit I need to send DEh (slave address) 08h (register address) and 41 (Enable WRTC bit, other bits remain in default). Why not hex? Why 41 and not 40? And what does SC22 in SC bit 6, SC21 in bit 5, etc. mean?
Datasheet
Example
I've read the documentation until I can't see anymore and I've searched until I am just getting more confused. Any help is appreciated.
Well, it looks like these values in the map are nibbles. The range for the first register is 0 - 59. When represented in BCD, 4 bits are needed for the digit in the ones place and three bits are needed for the 10's place. So, bits 0 - 3 belong to the first nibble; bit 0 = SC(register name)1(first nibble)0(first bit). Bits 4, 5 and 6 belong to the second nibble. Bit 4 = SC(register name)2(second nibble)0(first bit). Bit 7 is not needed.
The example sheet from Intersil has a typo; the WRTC value needs to be 40h or 41h.
I got this error on a CJ1W-CT021 card. It happen all of a sudden after its been running the program for some time. How i found it was by going to the IO Table and Unit Set up. Clicked on parameters for that card and found two settings in red.
Output Control Mode and And/Or Counter Output Patterns. This was there reading
Output Control Mode = 0x40 No Applicable Set Data
And/Or Counter Output Patterns = 0x64 No Applicable Set Data
no idea on how or why these would change they should of been
Output Control Mode = Range Mode
And/Or Counter Output Patterns = Logically Or
I have added some new code, but nothing big or really even used as i had the outputs of the new rungs jumped out. One thing i thought might cause this is every cycle of the program it was checking the value of an encoder connected to this card. Maybe checking it too offten? Anyhow if anyone has any idea what these do or how they would change please post.
Thanks
Glen
EDIT.. I wanted to add the bits i used, dont think any are part of this cards internal io but i may be wrong?
Work bits 66.01 - 66.06 , 60.02 - 60.07 , 160.12, 160.01 - 160.04, 161.02, 161.03
and
Data Bits (D)20720, 20500, 20600, 20000, 20590, 20040
I would check section 4-1 through 4-2-4 of the CT021 manual - make sure you aren't writing to reserved memory locations used for configuration data of the CT021 unit.
EDIT:
1) Check Page 26 of the above manual to see the location of the machine switch settings. The bottom dial sets the '1's digit and the top dial sets the '10's digit (ie machine number can be 0-99);
2) Per page 94, D-Memory is allocated from D20000 + (N X 100) (400 Words) where N is equal to the machine number.
I would guess that your machine number is set to 0 (ie: both dials at '0'), 5, or 6. In the case of machine number '0', this would make the reserved DM range D20000 -> D20399. In this case (see pages 97, 105) D20000 would contain configuration data for Output Control Mode (bits 00-07) and Counter Output Patterns (bits 08-15). It looks like you are writing 0x6440 to D20000 (or D20500, D20600 for machine number 5 or 6, respectively) and are corrupting the configuration data.
If your machine number is 0 then stay away from D20000-D20399 unless you are directly trying to modify the counter's configuration state (ie: don't use them in your program!).
If the machine number is 1 then likewise for D20100-D20499, etc. If you have multiple counters they can overlap ranges so they should always be set with machine numbers which are 4 apart from each other.
I am working on a QR code encoding/decoding project.
I have been read through the ISO/IEC 18004 (2006) and some tutorials ( http://www.thonky.com/guides/
http://www.matchadesign.com/_blog/Matcha_Design_Blog/post/QR_Code_Demystified_-_Part_1/
http://www.swetake.com/qr/qr1_en.html
)
The ISO documentation and those very nice tutorials helped me a lot. But there’s still one thing I can’t understand, that’s how we can calculate the number of data/error blocks when creating a QR code at Version 3 or higher.
The image below is from the ISO/IEC 18004 – 2006:
A version 7-H (H is error correction capacity level ) symbol that has 66 data codewords and 130 error codewords. They split both of them into 5 blocks.
The document says that the n blocks number (in this case n = 5 ) can be calculated from Table 9 (ISO 18004) according to the version and error correction level. But it seems like I can’t get that number. Please show me how I can calculate it.
Now I got it. All needed information for block splitting actually is at Table 9 of the ISO/IEC 18004 document. Just because of my careless reading.