anti aliasing a bresenham line

[falcore20]

Hi,

I am trying to anti-alias a bresenham line. I have this code so far. My only problem is that for the two variables that are declared as double, i have some problems. When i get to the code:

invD = 1.0 / (2.0 * Sqr(du * du + dv * dv))
invD2du = 2.0 * (du * invD)

It becomes

invD = 1# / (2# * Sqr(du * du + dv * dv))
invD2du = 2# * (du * invD)

There are # signs instead of decimals. Then i get overflow errors. Does anyone know how to fix this, or if there is a better way to anti alias the bresenham line? This would help me tremendously. Thankyou.

   dim uend As Integer
    Dim uincr As Integer
    Dim vincr As Integer
    Dim addr As Integer
    Dim incrS As Integer
    Dim incrD As Integer
    Dim twovdu As Integer
    
   Dim invD As Double
    Dim invD2du As Double

    Dim du As Integer
    Dim dv As Integer
    Dim u As Integer
    Dim v As Integer
    
    addr = (y0 * 640 + x0) * 4
    dx = x1 - x0
    dy = y1 - y0
    
    If Abs(dx) > Abs(dy) Then
    du = Abs(dx)
    dv = Abs(dy)
    u = x1
    v = y1
    uincr = 4
    vincr = 640 * 4
    If dx < 0 Then
        uincr = -uincr
    End If
        
    If dy < 0 Then
        vincr = -vincr
    End If
    
    Else
        
        du = Abs(dy)
        dv = Abs(dx)
        u = y1
        v = x1
        uincr = 640 * 4
        vincr = 4
        If dy < 0 Then
            uincr = -uincr
        End If
        
        If dx < 0 Then
            vincr = -vincr
        End If
    End If
    
    uend = u + 2 * du
    d = (2 * dv) - du
    incrS = 2 * dv
    incrD = 2 * (dv - du)
    twovdu = 0
    invD = 1.0 / (2.0 * Sqr(du * du + dv * dv))
    invD2du = 2.0 * (du * invD)
    
    Do
    
        PSet (addr, twovdu * invD)
        PSet (addr + vincr, invD2du - twovdu * invD)
        PSet (addr - vincr, invD2du + twovdu * invD)
        
        If d < 0 Then
            twovdu = d + du
            d = d + incrS
        Else
            twovdu = d - du
            d = d + incrD
            v = v + 1
            addr = addr + vincr
        End If
        u = u + 1
        addr = addr + uincr
    Loop While (u < uend)

[dafhi]

the # means the computer is recognizing the number as a double. if you put 1.1, the # would disappear, however it is best to leave it because to my knowledge, any number without a type declaration following it is treated as a variant. 9! means single, 9# means double, 4& means Long, 4% means integer. The # most likely PREVENTS overflow, so the problem is that your calculations are going way over.