Hi Dave,
You just use the decimal factor, if you want to move it 3 places the use
times 1000.
Now to keep things in the same format make sure use use the decimal point
all the time. Even though VB does not care, too loose in it's numbering systems.
123.4567890 * 1000000.0
Or a shift left would also do the same thing...
Bruce
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.
