Simplifying... (Arrays and Mod help) - Xtreme Visual Basic Talk

01-16-2004, 10:08 PM

spamonkey8 spamonkey8 is offline

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Simplifying... (Arrays and Mod help)

I have a little bit of code that I'd like to simplify in a project about permutations. It's a loop that, i'm sure, can be replaced by just the right Mod statement. There are two arrays, one large and one much smaller. What the loop does is for every byte in the larger array, the smaller is incremented one. When the position reaches the upper boundary of the smaller array, it goes back to zero. This is it:

Code:

For i = 1 To UBound(bytLarge())
    If lPosition = UBound(bytSmall()) Then
        lPosition = 0
    Else
        lPosition = lPosition + 1
    End If
Next
'I messed around with it and I found a statement that worked for many
'array dimensions, but it fails for some values, such as if the upper
'bounds are 41 and 5 for bytLarge and bytSmall, respectively
lPosition = ((UBound(bytLarge()) + 1) Mod (UBound(bytSmall()) + 1)) - 1

Thanks everybody

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01-16-2004, 10:30 PM

passel

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If the array will start at 0 then this should work

Code:

For i = 1 to UBound(bytLarge)
  IPosition = (IPosition + 1) Mod UBound(bytSmall)
Next

But I prefer to not repeatedly call functions if I can help it so I would
probably use

Code:

Dim L as long
L = UBound(bytSmall)
For i = 1 to UBound(bytLarge)
  IPosition = (IPosition + 1) Mod L
Next

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01-16-2004, 10:41 PM

spamonkey8 spamonkey8 is offline

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Wait, there's no way to use a single Mod call to do this? It's just a simple remainder problem, I don't see why I'd have to still loop.

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01-16-2004, 10:52 PM

passel

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I guess I don't understand what you're trying to do. ( I didn't quite
understand it the first time, but from your code, I thought my code
simplified what it was doing).
So

, what is it you're trying to do?

If you only want to get the same value for lposition with a mod instruction,
as you do when you run it through that loop and exit, then I guess you want

lposition = Ubound(bytLarge) Mod (UBound(bytSmall) + 1)

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Last edited by passel; 01-16-2004 at 11:15 PM.

01-16-2004, 11:12 PM

spamonkey8 spamonkey8 is offline

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Thank you!

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Last edited by spamonkey8; 01-16-2004 at 11:32 PM.