long integer conversion in VB3

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.

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

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.

If your four bytes are not numeric, how do you want to convert them in long?

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.

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

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

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.