Convert VB Double to Pascal Real

[David Scemama]

Hi,

I'm writing a program using VB.NET that needs to communicate with a DOS
Pascal program than cannot be modified. The communication channel is through
some file databases, and I have a huge problem writing VB Double values to
the file so as the Pascal program can read them as Pascal Real values.

I've managed to find the algorithm to read the Pascal Real format and
convert it to a VB Double, but I cannot figure out the opposite algorithm.

Can someone help me reverse my algorithm and develop the function
"DoubleToReal (ByVal Data As Double) As String"

Here is the conversion from real to double:

Public Function RealToDouble(ByVal Data As String) As Double
Dim dMantissa As Double
Dim i As Integer
Dim j As Long
Dim k As Long

If Len(Data) <> 6 Then
'Err.Raise
'exception
Exit Function
End If

'accumulate the mantissa
dMantissa = 1
For i = 6 To 2 Step -1
For j = CType(IIf(i = 6, 6, 7), Long) To 0 Step -1
k = k + 1
If (Asc(Mid$(Data, i, 1)) And CType(2 ^ j, Long)) <> 0 Then
dMantissa = dMantissa + 2 ^ -k
End If
Next j
Next i

'finally, assemble all the pieces into a number
If (Asc(Mid$(Data, 6, 1)) And &H80) = &H80 Then
RealToDouble = -dMantissa * 2 ^ (Asc(Mid$(Data, 1, 1)) - 129)
Else
RealToDouble = dMantissa * 2 ^ (Asc(Mid$(Data, 1, 1)) - 129)
End If

Try
Return ([Decimal].Round(CDec(RealToDouble), 2))
Catch ex As Exception
'MsgBox("RealToDouble, Conversion error: " & Data & "; " &
RealToDouble.ToString, MsgBoxStyle.Critical)
Return 0
End Try

End Function

Thanks for your help
David

 

[One Handed Man]

I probably have not understood you correctly, but can you not simply convert
the double to a string an then truncate the string to the correct length?

As far as I am aware, a real number is simply a number which is not
imaginary, so I dont know how this differs in a Pascal Real number. Your
function seems to convert a string into a double but goes a long way around.
BTW, I actually tried to convert "1.2345" and it returned 0.0

I'm sure Im missing something here, can you illuminate ?

Regards - OHM

 

[Cor]

Hi OHM

Can you take a look at that answer from cc I have sended in and give some
comments.

All is there just paste it on a new form and run.

Or I don't understand it (and no user will understand it) or it is a real
big bug.

Cor

 

[One Handed Man]

I think my reader must be out of sync here. I dont see this reply. I have
noticed this with quite a few posts lately. Can you repost under this one
please.

Thanks - OHM

 

[Cor]

Hi OHM this was a question from cc

how do I make keyboard & mouse temporary not response to desktop

I did want to answer with simple" me.enabled = false", but did not trust it
because this answer seems to simple to me. So I did test it, and I find
this strange and do not understand it.

Who will test it also and tell me if this is normal behaviour or a bug?
The problem is that when you pusth the button when enabled is false, it
keeps responding to the events, so you have to push several times on the not
locked button.

Framework 1.1

Cor

\\\
' needs one button and one label on a form
Private Sub Button1_Click(ByVal sender As System.Object, _
ByVal e As System.EventArgs) Handles Button1.Click
Static i As Integer
Me.Button1.Text = "locked"
Dim ctr As Control
' Me.Enabled = false did not work so
'I did try this but does not work either
For Each ctr In Me.Controls
ctr.Enabled = False
Next
Me.Refresh()
Dim y As Integer
For y = 1 To 20
Threading.Thread.Sleep(50)
i = i + 50
Me.Label1.Text = i.ToString
Me.Refresh()
Next
For Each ctr In Me.Controls
ctr.Enabled = True
Next
Me.Button1.Text = "Unlocked"
End Sub
///

 

[One Handed Man]

But this post is about Mouse and keyboard, the current thread we are in is
regarding Double/Pascal Real number conversions?

Whats happening, am I going mad ?

Regards OHM

 

[Cor]

Hi OHM

No I asket you to look to a posible Bug I had posted.

I was curious if I became crazy or it was a bug.

Cor

 

[David Scemama]

The Pascal Real type, is stored on 6 bytes with a very special coding. When
you write a real value in a file, 6 bytes are written to represent the value
(obviously, no language writes the string representation of the real !). My
function reads the 6 bytes and convert them to a VB double.

I need the reverse function !

David

I probably have not understood you correctly, but can you not simply convert
the double to a string an then truncate the string to the correct length?

As far as I am aware, a real number is simply a number which is not
imaginary, so I dont know how this differs in a Pascal Real number. Your
function seems to convert a string into a double but goes a long way around.
BTW, I actually tried to convert "1.2345" and it returned 0.0

I'm sure Im missing something here, can you illuminate ?

Regards - OHM

Hi,

I'm writing a program using VB.NET that needs to communicate with a
DOS Pascal program than cannot be modified. The communication channel
is through some file databases, and I have a huge problem writing VB
Double values to the file so as the Pascal program can read them as
Pascal Real values.

I've managed to find the algorithm to read the Pascal Real format and
convert it to a VB Double, but I cannot figure out the opposite
algorithm.

Can someone help me reverse my algorithm and develop the function
"DoubleToReal (ByVal Data As Double) As String"

Here is the conversion from real to double:

Public Function RealToDouble(ByVal Data As String) As Double
Dim dMantissa As Double
Dim i As Integer
Dim j As Long
Dim k As Long

If Len(Data) <> 6 Then
'Err.Raise
'exception
Exit Function
End If

'accumulate the mantissa
dMantissa = 1
For i = 6 To 2 Step -1
For j = CType(IIf(i = 6, 6, 7), Long) To 0 Step -1
k = k + 1
If (Asc(Mid$(Data, i, 1)) And CType(2 ^ j, Long)) <>
0 Then dMantissa = dMantissa + 2 ^ -k
End If
Next j
Next i

'finally, assemble all the pieces into a number
If (Asc(Mid$(Data, 6, 1)) And &H80) = &H80 Then
RealToDouble = -dMantissa * 2 ^ (Asc(Mid$(Data, 1, 1)) -
129) Else
RealToDouble = dMantissa * 2 ^ (Asc(Mid$(Data, 1, 1)) -
129) End If

Try
Return ([Decimal].Round(CDec(RealToDouble), 2))
Catch ex As Exception
'MsgBox("RealToDouble, Conversion error: " & Data & "; " &
RealToDouble.ToString, MsgBoxStyle.Critical)
Return 0
End Try

End Function

Thanks for your help
David

 

[One Handed Man]

This may be because the message goes out in the message queue, the button
does not respond to it until the loop finishes, then it processes the click
( I think ).

Regards - OHM

 

[Cor]

But should it do that?

 

[One Handed Man]

I dont know the answer to this. I suggest you post the following this
specific question at top level in the group:


Regards - OHM

 

[Cor]

Hi David,

I think this is a much to dificult approach for a newsgroup.
Maybe if guys like Fergus Cooney where here now, he could help you because
he likes this.

But he is not and I don't know if he will return soon.

Every byte in a file is representent by a hex value.

Can you not represent the bytes in hex form representing a normal decimal
value, maybe somebody can help you than.

Probably not me, because mostly this is not my stuff, but you never know.

Cor

 

[Cor]

Hi OHM,

Putted it in the vb.language.controls group.

Almost the private group from Herfied, I am curious what happens.

)

Cor

 

[David Scemama]

The exact representation of the Pascal Real type is:
Sign Significand Exponent
Width (bits) 1 39 8

David

The Pascal Real type, is stored on 6 bytes with a very special coding. When
you write a real value in a file, 6 bytes are written to represent the value
(obviously, no language writes the string representation of the real !). My
function reads the 6 bytes and convert them to a VB double.

I need the reverse function !

David

I probably have not understood you correctly, but can you not simply convert
the double to a string an then truncate the string to the correct length?

As far as I am aware, a real number is simply a number which is not
imaginary, so I dont know how this differs in a Pascal Real number. Your
function seems to convert a string into a double but goes a long way around.
BTW, I actually tried to convert "1.2345" and it returned 0.0

I'm sure Im missing something here, can you illuminate ?

Regards - OHM

Hi,

I'm writing a program using VB.NET that needs to communicate with a
DOS Pascal program than cannot be modified. The communication channel
is through some file databases, and I have a huge problem writing VB
Double values to the file so as the Pascal program can read them as
Pascal Real values.

I've managed to find the algorithm to read the Pascal Real format and
convert it to a VB Double, but I cannot figure out the opposite
algorithm.

Can someone help me reverse my algorithm and develop the function
"DoubleToReal (ByVal Data As Double) As String"

Here is the conversion from real to double:

Public Function RealToDouble(ByVal Data As String) As Double
Dim dMantissa As Double
Dim i As Integer
Dim j As Long
Dim k As Long

If Len(Data) <> 6 Then
'Err.Raise
'exception
Exit Function
End If

'accumulate the mantissa
dMantissa = 1
For i = 6 To 2 Step -1
For j = CType(IIf(i = 6, 6, 7), Long) To 0 Step -1
k = k + 1
If (Asc(Mid$(Data, i, 1)) And CType(2 ^ j, Long)) <>
0 Then dMantissa = dMantissa + 2 ^ -k
End If
Next j
Next i

'finally, assemble all the pieces into a number
If (Asc(Mid$(Data, 6, 1)) And &H80) = &H80 Then
RealToDouble = -dMantissa * 2 ^ (Asc(Mid$(Data, 1, 1)) -
129) Else
RealToDouble = dMantissa * 2 ^ (Asc(Mid$(Data, 1, 1)) -
129) End If

Try
Return ([Decimal].Round(CDec(RealToDouble), 2))
Catch ex As Exception
'MsgBox("RealToDouble, Conversion error: " & Data & "; " &
RealToDouble.ToString, MsgBoxStyle.Critical)
Return 0
End Try

End Function

Thanks for your help
David

 

[David Scemama]

The exact representation of the Pascal Real type is:
Sign Significand Exponent
Width (bits) 1 39 8

David

The Pascal Real type, is stored on 6 bytes with a very special coding. When
you write a real value in a file, 6 bytes are written to represent the value
(obviously, no language writes the string representation of the real !). My
function reads the 6 bytes and convert them to a VB double.

I need the reverse function !

David

I probably have not understood you correctly, but can you not simply convert
the double to a string an then truncate the string to the correct length?

As far as I am aware, a real number is simply a number which is not
imaginary, so I dont know how this differs in a Pascal Real number. Your
function seems to convert a string into a double but goes a long way around.
BTW, I actually tried to convert "1.2345" and it returned 0.0

I'm sure Im missing something here, can you illuminate ?

Regards - OHM

Hi,

I'm writing a program using VB.NET that needs to communicate with a
DOS Pascal program than cannot be modified. The communication channel
is through some file databases, and I have a huge problem writing VB
Double values to the file so as the Pascal program can read them as
Pascal Real values.

I've managed to find the algorithm to read the Pascal Real format and
convert it to a VB Double, but I cannot figure out the opposite
algorithm.

Can someone help me reverse my algorithm and develop the function
"DoubleToReal (ByVal Data As Double) As String"

Here is the conversion from real to double:

Public Function RealToDouble(ByVal Data As String) As Double
Dim dMantissa As Double
Dim i As Integer
Dim j As Long
Dim k As Long

If Len(Data) <> 6 Then
'Err.Raise
'exception
Exit Function
End If

'accumulate the mantissa
dMantissa = 1
For i = 6 To 2 Step -1
For j = CType(IIf(i = 6, 6, 7), Long) To 0 Step -1
k = k + 1
If (Asc(Mid$(Data, i, 1)) And CType(2 ^ j, Long)) <>
0 Then dMantissa = dMantissa + 2 ^ -k
End If
Next j
Next i

'finally, assemble all the pieces into a number
If (Asc(Mid$(Data, 6, 1)) And &H80) = &H80 Then
RealToDouble = -dMantissa * 2 ^ (Asc(Mid$(Data, 1, 1)) -
129) Else
RealToDouble = dMantissa * 2 ^ (Asc(Mid$(Data, 1, 1)) -
129) End If

Try
Return ([Decimal].Round(CDec(RealToDouble), 2))
Catch ex As Exception
'MsgBox("RealToDouble, Conversion error: " & Data & "; " &
RealToDouble.ToString, MsgBoxStyle.Critical)
Return 0
End Try

End Function

Thanks for your help
David

 

[One Handed Man]

You'll probably get the reply in German !.

Regards - OHM

 

[One Handed Man]

Hay David,

This is clearly a complex conversion. I found this on the internet. It is
Quick Basic rendering of a solution to your problem ( Hopefully ). You will
need to amend it a bit for VB.NET, but hopefully it gives you a template to
work with.

HTH - OHM


DECLARE FUNCTION power# (x!, y AS INTEGER)
DECLARE SUB RealConv (RealCost AS ANY, NewCost#)
' the QBASIC equivalent of the above Pascal struct:
TYPE PASdataRecord
EmpNameLength AS STRING * 1
EmpName AS STRING * 30
Number AS INTEGER
Wage AS STRING * 6
END TYPE
' set up the file to be opened and read from QBASIC
DIM EmployeeDAT AS PASdataRecord
OPEN "EMPLOY.DAT" FOR RANDOM ACCESS READ LOCK WRITE AS #1 LEN =
LEN(EmployeeDAT)
' read the file a record at a time until the end
DO WHILE NOT EOF(1)
CLS
Count = Count + 1
SEEK #1, Count
GET #1, , EmployeeDAT
' strip out the actual string using the first length byte
EmployeeDAT.EmpName = MID$(EmployeeDAT.EmpName, 1,
ASC(EmployeeDAT.EmpNameLength))
' the following routine converts the pascal real to Qbasic double
CALL RealConv(EmployeeDAT.Wage, BASwage#)
PRINT "Employee number = "; EmployeeDAT.Number
PRINT "Employee EmpName = "; EmployeeDAT.EmpName
PRINT "Employee wage = "; BASwage
LOOP


FUNCTION power# (x, y AS INTEGER)
' simple x to the power of y function
power# = EXP(y * LOG(x))
END FUNCTION

SUB RealConv (Real$, NewCost#)
' create an array to hold each byte of the real string
DIM RealHold(6)
RealHold(1) = ASC(MID$(Real$, 1, 1))
RealHold(2) = ASC(MID$(Real$, 2, 1))
RealHold(3) = ASC(MID$(Real$, 3, 1))
RealHold(4) = ASC(MID$(Real$, 4, 1))
RealHold(5) = ASC(MID$(Real$, 5, 1))
RealHold(6) = ASC(MID$(Real$, 6, 1))
' if positive contains a number then its negative
positive = &H80 AND RealHold(6)
' clear the Pos/Neg bit from byte 6
RealHold(6) = &H80 OR RealHold(6)
' set up the significand as 1.0
Significand# = 1#
' check each individual bit for on/off; if on then multiply out the
' number (2,4,8,16,32,64,128, etc.)
FOR bytecheck = 2 TO 6
' bit 0 of byte
IF (RealHold(bytecheck) AND &H1) = 1 THEN
Significand# = Significand# + power(2, (0 + (bytecheck - 2) * 8))
END IF
' bit 1 of byte
IF (RealHold(bytecheck) AND &H2) = 2 THEN
Significand# = Significand# + power(2, (1 + (bytecheck - 2) * 8))
END IF
' bit 2 of byte
IF (RealHold(bytecheck) AND &H4) = 4 THEN
Significand# = Significand# + power(2, (2 + (bytecheck - 2) * 8))
END IF
' bit 3 of byte
IF (RealHold(bytecheck) AND &H8) = 8 THEN
Significand# = Significand# + power(2, (3 + (bytecheck - 2) * 8))
END IF
' bit 4 of byte
IF (RealHold(bytecheck) AND &H10) = 16 THEN
Significand# = Significand# + power(2, (4 + (bytecheck - 2) * 8))
END IF
' bit 5 of byte
IF (RealHold(bytecheck) AND &H20) = 32 THEN
Significand# = Significand# + power(2, (5 + (bytecheck - 2) * 8))
END IF
' bit 6 of byte
IF (RealHold(bytecheck) AND &H40) = 64 THEN
Significand# = Significand# + power(2, (6 + (bytecheck - 2) * 8))
END IF
' bit 7 of byte
IF (RealHold(bytecheck) AND &H80) = 128 THEN
Significand# = Significand# + power(2, (7 + (bytecheck - 2) * 8))
END IF
NEXT
' normalize the number by dividing calculated number by a number with all
' bits turned on: 2 to the power of 40
Significand# = Significand# / power(2, 40)
' calculate in the exponent
Number# = Significand# * power(2, (RealHold(1) - 128))
' set the pos/neg sign
IF positive > 0 THEN Number# = Number# * -1
NewCost# = Number#
END

 

[David Scemama]

Unfortunately, this is the exact contrary of what I would like to do. This
function converts the Pascal Real contained in a string to a double.

Thanks
David