Hi David,
Doug has given you the answer, but just to try to add more to it: binary floating point numbers almost never are stored in a form which exactly equals a string you can type in; even when you enter 123.456789 into a variable, the contents of this variable are not stored so that they can be relied on to exactly equal 123.456789. Sometimes it will, sometimes it won't. This is why, the results of most decimal calculations are rounded to some expected number of decimal digits to the right of the decimal point when displayed. If you need this decimal value as a result of working with user input, or something stored in a database, as Doug says, the only way you can rely on reproducing what the user entered is to store the information as a string, or, make use of the round function to display a known (and limited) number of characters to the right of the decimal point. I come up against this in the routines where I work with latitude and longitude coordinates stored in decimal degree formats. I want to do numeric calculations on them, so whenever I show them to the user I have to use the round function to limit the number of characters shown. Other programming languages do offer binary decimal formats which exactly represent a defined number of decimals (and VBScript does when you use the currency type), but VBScript doesn't offer this; just the binary floating point. Hth, Chip From: David [mailto:[email protected]] Sent: Tuesday, January 01, 2013 9:27 AM To: [email protected] Subject: VBS code - floating point numbers I have been playing for a while now, with a tricky numeric issue. Hopefully, it is just that I have overlooked some kind of instruction. What I am trying to do, is to derive the decimals, from a floating point number. Let's for instance say, you have the number: 123.4567890 I know, you can use the Int Or CInt instructions on this number, and it will return 123. That is, the digits to the left of the decimal point. But what I am after, is a way to have returned the digits to the right of the decimal point - in the example above, that would give 4567890. I really can't seem to find an instruction that will let me do this. Or, maybe it is named somthing, that my English knowledge would not have included. Smile. So if anyone out there, would happen to know of such an instruction or workarounds for this task, I would greatly appreciate your feedback. OK, You would think, that if we did a basic piece of math, things should not get too complicated. So I thought, if I take my original number, and subtract the Int value of the number, I would end up with a simple decimal number, that I then could do some extra work on. Well, I tried a code like this: X = 123.4567890 speak x -int(x) . What I did expect, was to get the return value of "0.4567890". But, that's not what I get. My script will speak out the number of 0.456789000000001 . and if you do a more direct way, like: speak 123.4567890 -123 ; you might end up with an even less predictable number. Well, my idea was - for a workaround - to have converted the returned value into a string, and then simply omitted the first two characters (which would be 0.). Then, I could have converted the final string back to a number, and had the job done. Not exactly anything straight forward, but it would have been a workaround. Yet, long as the returned value is not as expected, that workaround would not be useful. Again, is there an instruction in VBS, that will directly return only the digits to the right of the decimal point? One more thing. Some of you, might be wondering, why don't I just convert the original number into a string, and then split by the decimal point. A code like this: x = 123.4567890 NumString = Split(X, ".") Speak NumString(1) . OK, this would work perfectly; long as we are operating numbers that follow English standard, with the dot-sign as the decimal point. But in other languages, the comma-sign is used as a decimal point. Since I want the script to work, no matter the locale setting of the computer, and the corresponding decimal point character, I would have been more satisfied with a direct instruction for deriving the digits to the right of the decimal point. Hope all of this makes sense, and that someone could give me a kick in the right direction for a solution.
