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Can I read rearranged nibbles across bytes as a single value from lua

asked 2020-09-10 08:55:32 +0000

Mayuri gravatar image

The wireshark capture is having distorted data across bytes. I have rearranged it by nibbles in Lua script, but not getting how can i read the rearranged nibbles as a single value in order to obtain correct value.

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Can you add an illustration of what you mean, e.g.

| Lo Nibble | Hi Nibble | to make a byte or
| LoLo Nibble | Lo Nibble | Hi Nibble | HiHi Nibble | to make a 16 bit value

or something else?

You could accumulate the nibbles by multiplying, e.g. local byte = LoNibble + HiNibble * 16.

grahamb gravatar imagegrahamb ( 2020-09-10 09:46:25 +0000 )edit

Below is the capture: 00000000001000001000000000000101 = 0x00 0x20 0x80 0x05 Want to read a value as: 00000000001000000000000001011000 = 0 512 88

So, is there a way to rearrange it and then read it?

Mayuri gravatar imageMayuri ( 2020-09-10 11:53:45 +0000 )edit

Not sure I follow, you have:

|0000|0000|0010|0000|1000|0000|0000|0101|
|0x0 |0x0 |0x2 |0x0 |0x8 |0x0 |0x0 |0x5 |
| #0 | #1 | #2 | #3 | #4 | #5 | #6 | #7 |

and you want it to be:

|0000|0000|0010|0000|0000|0000|0101|1000
|0x0 |0x0 |0x2 |0x0 |0x0 |0x0 |0x5 |0x8 |
| #0 | #1 | #2 | #3 | #6 | #5 | #7 | #4 |

How are the nibbles rearranged there? You seem to have swapped the last two nibbles and then 4th and the last.

grahamb gravatar imagegrahamb ( 2020-09-10 13:16:40 +0000 )edit

There would seem to be a typo as 0 512 88 would be 0x00 0x02 0x00 0x58 and not 0x00 0x20 0x00 0x58. Is that what was intended?

cmaynard gravatar imagecmaynard ( 2020-09-11 04:16:00 +0000 )edit

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answered 2020-09-11 04:53:37 +0000

cmaynard gravatar image

Judging by the "Want to read a value as ..." comment where 0 512 88 was desired, it looks to me like you have 3 different fields split across 4 bytes. The only practical way I can see how to get the desired numerical results from the original data is to manipulate the bytes as follows:

  • Rotate right (with wrap) the 1st 2 bytes
  • Rotate left (with wrap) the 2nd 2 bytes
  • Combine the data
  • Isolate the 3 individual fields from the combined result

It may be possible to optimize this, but here’s an implementation based on the above assumptions:

    -- Rotate the 1st 2 bytes to the right and the 2nd 2 bytes to the left
    local x = 0x00208005
    local x_hi = bit.rshift(bit.band(x, 0xffff0000), 16)
    local x_lo = bit.band(x, 0x0000ffff)
    local x_hi_ror = bit.bor(bit.ror(x_hi, 4), bit.lshift(bit.band(x_hi, 0x000f), 12))
    local x_lo_rol = bit.bor(bit.band(bit.rol(x_lo, 4), 0x0fff), bit.rshift(x_lo, 12))
    local x_new = bit.bor(bit.lshift(x_hi_ror, 16), x_lo_rol)

    print("x = " .. string.format("0x%04x", x) .. "\n" ..
        "x[1] = " .. bit.rshift(bit.band(x, 0xff000000), 24) .. "\n" ..
        "x[2] = " .. bit.rshift(bit.band(x, 0x00ffff00), 8) .. "\n" ..
        "x[3] = " .. bit.band(x, 0x000000ff))
    print("x_new = " .. string.format("0x%04x", x_new) .. "\n" ..
        "x_new[1] = " .. bit.rshift(bit.band(x_new, 0xff000000), 24) .. "\n" ..
        "x_new[2] = " .. bit.rshift(bit.band(x_new, 0x00ffff00), 8) .. "\n" ..
        "x_new[3] = " .. bit.band(x_new, 0x000000ff))

Refer to the Lua Bit Operations Module for more information on the bit operations used in this example.

NOTE: If this answer is based on faulty assumptions and doesn't help solve your problem, then you'll have to provide more information.

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Asked: 2020-09-10 08:55:32 +0000

Seen: 102 times

Last updated: Sep 11