long integer conversion in VB3
How do I extract a long integer from a file (stored using MKL$ in MsBasic 7,.2) and store the same long back to the file.
I've tried extrecting it with :
dim a as integer
dim data as string * 6
a& = Clng(left(data$,4))
but it returns Type Mismatch
I've tried extrecting it with :
dim a as integer
dim data as string * 6
a& = Clng(left(data$,4))
but it returns Type Mismatch
You should declare the variable 'a' as long, and you have to care that all first 4 characters in 'data' are numeric.
dim a as integer
dim data as string * 6
a& = Clng(left(data$,4))
To save the variable back to a file, you could use the Put statment.
dim a as integer
dim data as string * 6
a& = Clng(left(data$,4))
To save the variable back to a file, you could use the Put statment.
I was just preparing this to answer the question, I found it was already answered but I thought I would post this anyway
Option Explicit
Private Sub Form_Load()
Dim a As Long
Dim data As String * 6
'My test data
data = Chr$(&H7) & Chr$(&H5B) & Chr$(&HCD) & Chr$(&H15)
a = nConvToLong(Left$(data, 4))
MsgBox a
End Sub
Private Function nConvToLong(ByVal sInp As String) As Long
Dim nRet As Long
Dim i As Integer
nRet = 0
If Len(sInp) > 0 Then
For i = 1 To Len(sInp)
nRet = nRet * 256
nRet = nRet + Asc(Mid$(sInp, i, 1))
' MsgBox Hex$(nRet)
Next i
End If
nConvToLong = nRet
End Function
Option Explicit
Private Sub Form_Load()
Dim a As Long
Dim data As String * 6
'My test data
data = Chr$(&H7) & Chr$(&H5B) & Chr$(&HCD) & Chr$(&H15)
a = nConvToLong(Left$(data, 4))
MsgBox a
End Sub
Private Function nConvToLong(ByVal sInp As String) As Long
Dim nRet As Long
Dim i As Integer
nRet = 0
If Len(sInp) > 0 Then
For i = 1 To Len(sInp)
nRet = nRet * 256
nRet = nRet + Asc(Mid$(sInp, i, 1))
' MsgBox Hex$(nRet)
Next i
End If
nConvToLong = nRet
End Function
Dear Waty,
Sorry I meant to declare it as a Long Integer but it still won't work. As it is a Long, the 4 Bytes are therefore not numeric
Dear Swilt
Lovely bit of coding but I can't get it to work.
Sorry I meant to declare it as a Long Integer but it still won't work. As it is a Long, the 4 Bytes are therefore not numeric
Dear Swilt
Lovely bit of coding but I can't get it to work.
Dear Waty,
In MSBasic an MKL$ converts a numeric to a long i.e. four bytes
What I have now discovered, with the aid of swilt, is that is is held as 4 bytes in reverse, so that reversing the original 4 byte s (byte 4,3,2 and 1 to byte 1,2,3 and 4) and taking the ascii value and multiplying by 256 give the converted number.
In MSBasic an MKL$ converts a numeric to a long i.e. four bytes
What I have now discovered, with the aid of swilt, is that is is held as 4 bytes in reverse, so that reversing the original 4 byte s (byte 4,3,2 and 1 to byte 1,2,3 and 4) and taking the ascii value and multiplying by 256 give the converted number.
Dear Swilt,
You got answer right but didn't explain that I had to reverse the original Long Integer. So I think you should get the points but I don't know how to release them.
Watch out for the same questions but the singles and doubles.
Many Thanks
You got answer right but didn't explain that I had to reverse the original Long Integer. So I think you should get the points but I don't know how to release them.
Watch out for the same questions but the singles and doubles.
Many Thanks
p.s.
Use For i = Len(sInp) to 1 step -1 if the number is round the other way and watch out for negative numbers because I have not tested for them
Use For i = Len(sInp) to 1 step -1 if the number is round the other way and watch out for negative numbers because I have not tested for them
A long is just four bytes as a character string. Here's an old inline function that I wrote in Quick Basic to work with a long value that I had read in as four bytes:
DEF FNS2L& (s$)
hex2& = ASC(MID$(s$, 2, 1))
hex2& = hex2& * 256
hex3& = ASC(MID$(s$, 4, 1))
hex3& = hex3& * 4096
hex4& = ASC(MID$(s$, 3, 1))
hex4& = hex4& * 65536
FNS2L& = (ASC(MID$(s$, 1, 1))) + hex2& + hex3& + hex4&
END DEF
To use this you would read in the data into a four byte string variable. Using binary file i/o and assuming your integer is at the beginning of the file that would be simply:
open "FileName" for binary as #1
x$=space$(4)
get #1,,x$
close #1
and say you convert the above to a basic function:
lngvar&=FNS2l&(x$)
Under Quick Basic these strings were more easily handled through the availability of the MKL$, CVL instructions now missing from VB.
DEF FNS2L& (s$)
hex2& = ASC(MID$(s$, 2, 1))
hex2& = hex2& * 256
hex3& = ASC(MID$(s$, 4, 1))
hex3& = hex3& * 4096
hex4& = ASC(MID$(s$, 3, 1))
hex4& = hex4& * 65536
FNS2L& = (ASC(MID$(s$, 1, 1))) + hex2& + hex3& + hex4&
END DEF
To use this you would read in the data into a four byte string variable. Using binary file i/o and assuming your integer is at the beginning of the file that would be simply:
open "FileName" for binary as #1
x$=space$(4)
get #1,,x$
close #1
and say you convert the above to a basic function:
lngvar&=FNS2l&(x$)
Under Quick Basic these strings were more easily handled through the availability of the MKL$, CVL instructions now missing from VB.
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