Lost Digits of big Number - convert Double into Hex

Question

My Number is 24563383270390080
if i debug the Variable it has got a value of 2.45633832703901e+016, so i lost a little bit of the value.
if i bring ist into a sting, i can see the complete value "24563383270390080"
i have to convert the number in to a Hex-Value, the Reuslt should be "$57444456503941" an can be written into a string.
my problem is to bring the original value into a string of hex, because diadem round the last digits of my number
i tryed several convertions, but the the probmel of lost digits is still there....
maybe you´ve got an idea
dim Number, NumberString, HexTest1, HexTest2, HexTarget
Number = 24563383270390080  '17 Digits
HexTarget = "$57444456503941"
NumberString = Str(Number, "d")
HexTest1  = Str((Data.Root.ChannelGroups("Neu").Channels("Vin_Data").Values(1)), "$")
HexTest2  = Str(Val(Number), "$")
'---> Result
''--> Number       = 2.45633832703901e+016
'--> NumberString = "24563383270390080"
'--> HexTarget  = "$57444456503941"  
'--> HextTest1  =  $57444456503940  !!!!! Last Postion !!!!
'--> HextTest2  =  $57444456503940

Answer 1

You are a little bit out of luck with your example because
dim number : Number = 24563383270390080 '17 Digits
dim txt
txt = txt & Str(Number, "d") & " ref: 24563383270390080" & VBCRLF
txt = txt & Str(Number, "$") & " ref: $57444456503940" & VBCRLF
MsgBox txt
 0 at the end is correct. (At least the MS calculator says so.) But this is by incident because of the number you picked.
VBS only supports 64 bit double values. 64 bit doubles have a mantissa of 52 bits.
24563383270390080
becomes
1010111010001000100010001010110010100000011100101000000
which has 55 bit in use. Because your last bits aren't in use the number you picked can be represented by a double without loosing precission.
24563383270390081
would show the behavior you described because it is using 55bit range. The 1 is just cut when converted to double.
But I have no solution for this because it will not work with VBS.