Performing mathematical operations on byte arrays

Friends,

Could I have your valuable advice on how to do this in VB6. I am not familiar if VB6 can perform this kind of vector algebra by some preset commands.


I have 2 long byte arrays :

ByteArr_X = [0000000011111000000001111]
ByteArr_Y= [0010010011111000000001111]

I want to make a new byte array:

ByteArr_Z

such that:

first element of ByteArr_X* 1 + first element of ByteArr_Y *2
second element of ByteArr_X* 1 + second element of ByteArr_Y *2
third element of ByteArr_X* 1 + third element of ByteArr_Y *2
...
...
...
last element of ByteArr_X* 1 + last element of ByteArr_Y *2

Can VB6 perform vector mathematics, or how does one approach this type of mathematics in VB6?

Calculus

Are you using VB6?

Clarification: What are the contents of the Byte array? Are each element just a 1 or 0? Or do they have values from 0 to 255? By first element, do you simply mean the entire byte value or are you referencing bits vs bytes?

Dear LaVolpe,

First thanks for your help with my earlier problem.

Yes you are right,each element just a 1 or 0.

The first element of the first array is 0 , the second is also 0 and so on, sorry, I was using Matlab kind of terminology.

I have declared the byte arrays in the following manner:

Dim ByteArr_X() As Byte, ByteArr_Y() As Byte

So by first element I mean the first value.

Thanks

Calculus

Calculus

You're in great hands with LaVolpe, so I'll try not to step on his toes...
but, a general question:

how do you multiply ByteArr_Y by 2 and represent that as a byte?

eg:

baX(0) = 0, baY(0) = 0 >> 0*1 + 0*2 = 0 .. baZ(0) = 0
baX(1) = 0, baY(1) = 0 >> 0*1 + 0*2 = 0 .. baZ(1) = 0
baX(2) = 0, baY(2) = 1 >> 0*1 + 1*2 = 2 .. baZ(2) = 2 << huh ??
baX(3) = 0, baY(3) = 0 >> 0*1 + 0*2 = 0 .. baZ(3) = 0

Spoo

You're welcome, regarding previous assistance. We don't always get a thank you, but it makes our day when we do get them.

You do realize that adding 1 to 1 will be 2. Is that permitted? And if not, what happens to your array when that does occur.

If values of 2 are no biggie, then your loop might look like this

Code:

---

Dim ByteArr_Z() As Byte
Dim i As Long
ReDim ByteArr_Z(LBound(ByteArr_X) To UBound(ByteArr_X))
For i LBound(ByteArray_X) To UBound(ByteArr_X)
  ByteArr_Z(i) = ByteArray_X(i) + ByteArray_Y(i)
Next

---

Above loop pertains to your addition requirement

Quote:

---

Originally Posted by calculus

first element of ByteArr_X* 1 + first element of ByteArr_Y *2
second element of ByteArr_X* 1 + second element of ByteArr_Y *2
third element of ByteArr_X* 1 + third element of ByteArr_Y *2

---

Thanks Spoo and LaVolpe,

Firstly, Spoo yes it may seem strange but this is exactly the input required by the device I am trying to operate, basically it wants inputs as a single array of numbers , it then does backwards calculation and gets the binary form for each channel.
:)

LaVolpe, Thank you very much once again, I will try this out and get back to you in a few days.

You guys are great.

Calculus

Calculus

OK, glad the suggestions were deemed helpful. Two things...

1. I think my question stemmed from incorrect use of terms bit and byte.
Natch, a element of a byte array can range in value from 0 to 255, so
a value of 2 is not a problem (despite my initial balk).

2. I think LaVolpe omitted the multiplication component in his code..

Code:

---

Dim ByteArr_Z() As Byte
Dim i As Long
ReDim ByteArr_Z(LBound(ByteArr_X) To UBound(ByteArr_X))
For i LBound(ByteArray_X) To UBound(ByteArr_X)
  ByteArr_Z(i) = ByteArray_X(i) + ByteArray_Y(i) * 2
Next

---

.. but I'm sure you both already noticed that, too.

Look forward to your update in a few days.

Spoo

Quote:

---

Originally Posted by Spoo

2. I think LaVolpe omitted the multiplication component in his code.... but I'm sure you both already noticed that, too.

---

I didn't notice my blunder, hopefully calculus did. Thanx :o

:) Yes I did, thank you both.

By the way does VB6 allow for bit arrays, kind of curious because that will certainly save a lot of computer memory. If not its fine.

Thanks

Calculus

No bit arrays. Bits calculations must be performed via bit shifting and capable of handling overflows; there are plenty of posts on this topic and I wouldn't say it is always easy, but challenging would be a good word.

I see. Thanks LaVolpe