For those of you who don't understand non-Base10 numbers....

[Captain Pinko]

As we all no binary is a system of counting using only 0's and 1's. But how does this work? Well imagine an odometer (distance thingy in your car) that after it reaches a max number on one wheel it flips the one next to it one position. The base of the counting system denotes the number of positions on each. We use base 10 (ie. DECimal, DEC meaing 10) and there are 10 possible positions for our odometer 0,1,2,3,4,5,6,7,8,9. To convert a number from its base to base ten:
1) take a number
2) multiply its by its value
3) then by its base to the power of its' distance from the decimal point (NB the number to the left of the decimal point is distance 0, the one to the right is -1)
4) add it to a running total
5) do with all numbers

EXAMPLE USING HEXADECIMAL

number = A03D.C
1) A
2) A = 10
3) Base = 16
distance = 3
10 * 16^3 = 40960
4) Runing Total = 40960
1) 0
2) 0 = 0
3) base = 16
distance = 2
0 * 16^2 = 0
4) RT = 40960 + 0 = 40960
1) 3
2) 3 = 3
3) distance = 1
3 * 16^1 = 48
4) RT = 40960 + 48 = 41008
1) D
2) D = 13
3) distance = 1
13 * 16^0 = 13
4) RT = 356128 + 208 = 41021
1) C
2) C = 12
3) distance = -1
12 * 16^-1 = 0.75
4) RT = 41021.75
Therefore A03D.C = 41021.75
(check it on your WinCalculator)
you can see that it is neater and shorter.
for other bases replace 26 with the appropriate base #.
Please respond if you found this helpful or want clarification.

[kedaman]

Here's something you could have use for, you could convert from between whatever bases. The elements used beyond 9 continues on A-Z and [\] and so on

Function Base2Dec(val$, base%): Dim n%, a() As Byte
    a = StrConv(UCase(val), vbFromUnicode)
    For n = 1 To Len(val)
        Base2Dec = CDec(Base2Dec + (a(n - 1) - 48 + 7 * (a(n - 1) > 64)) * base ^ (Len(val) - n))
    Next n
End Function
Function Dec2Base$(val, base%): Dim n%, s
    For n = 0 To 100
        s = Int(val / base ^ n)
        If s = 0 Then Exit For
        s = s - Int(s / base) * base 's mod base
        Dec2Base = Chr(s + 48 - 7 * (s > 9)) & Dec2Base
    Next n
End Function

cool, I have been looking for code like that for a long time

[kb244]

I think it might be usefull to understand the base numbers in binary

1 - 1
2 - 10
4 - 100
8 - 1000
16- 10000
32- 100000
64- 1000000
128-10000000
256-100000000
512-1000000000
1024-10000000000

each binary increases to another digit
you add one to the right, you move it up until you get to the end for example (i'm only going to do the first few)

1
10
11
100
101
110
111
1000
1001
1010
1100
1101
1110
1111
10000

also keep note all the way upto 256 , hexidecimal remains 2 digit ( Octal stops at 8 )
hexidecimals are used mainly in representation, because the digits can remain small even if the numbers get big.

1 - 1 - 1
2 - 10 - 2
4 - 100 - 4
8 - 1000 - 8
16 - 10000 - 10
32 - 100000 - 20
64 - 1000000 - 40
128 - 10000000 - 80
256 - 100000000 - 100 ( 255 is FF, numerically it goes from 0-FF for 0-255, which still makes a count of 256)

see the patern in each base?

up til 255 Hex keeps a consitant 2 digit representation.
so for each binary step it adds a digit, for each hexidecimal step, it doubles itself(or would seem like it doubles itself if you looked at it from a 10base perspective)

but the source codes above are good if you just want to get the representation conversion without understanding it all (btw I did not use a calculator or anything, it's from memory, if you want to check my work go ahead)

[Captain Pinko]

Thanx for replying but the point was to explain how they work. And as for memorizing the binary it is not all that hard. Just take number that is two to the power of XXXXX and stick a 1 in that spot (eg. 512 = 2^9 so = 100000000 in binary)