That certainly works for variable, but it breaks down with objects. I
expanded my sample test and these are the results:
---
OOPic Code:
Dim firstByte As Byte
Dim secondByte As Byte
Dim thirdByte As Byte
Dim firstWord As Word
Dim secondWord As Word
Dim firstObj As oByte
Dim secondObj As oByte
Dim thirdObj As oMathO
Sub Main()
..firstByte = 15 '0F hex
..secondByte = 171 'AB hex
..thirdByte = 205 'CD hex
..firstWord = 4660 '1234 hex
..secondWord = 43981 'ABCD hex
..firstObj = 15
..secondObj = 1
..thirdObj.Input1.Link(firstObj)
..thirdObj.Input2.Link(secondObj)
..thirdObj.Mode = cvAdd
..thirdObj.Operate = cvOn
End Sub
---
Mapping in information box in the OOPic interface:
firstByte = Bank1, AreaV, Bank in sub address: 2
secondByte = Bank1, AreaV, Bank in sub address: 2
thirdByte = Bank1, AreaV, Bank in sub address: 2
firstObj = Bank0, AreaI, Bank in sub address: 0
secondObj = Bank0, AreaI, Bank in sub address: 0
thirdObj = Bank0, AreaA, Bank in sub address: 0
Note that we would expect that firstObj and secondObj would use bank 1
in the sub address to be consistant with the "areas map to
banks" theory, but it isn't the same in practice.
---
Lines from the Omp file:
V.I-126.....2..oByte.........firstObj
V.I-124.....2..oByte.........secondObj
V.A-41......5..oMathO........thirdObj
---
The conversation:
Send: \0V
Recv: v
Send: U
Send: 0H126J0LM = read value of firstObj
Recv: 15m = expected value
Send: 0H41J0LM = read value of thirdObj
Recv: 16m = expected value
This test would imply that things in Area A, and Area I need a sub
address bank of 0, whereas things in Area V need a sub address bank of
2. I have no idea what Area F is for.
Ian, if you have time would you independently verify these findings. I
need a sanity check.
Thanks,
Dogulas
[Non-text portions of this message have been removed]
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