If "%f" with no modifiers is passed to a standard sprintf function, it
places 6 digits after the decimal point. So the values that sprintf
produced for me were (using Unix):
12.345000
123.400002
1234.560059
1234.500000
12345.000000
Some rounding errors were also introduced.
I had a case about a month ago where I needed to print numbers in floating
point format, and did not want the scientific format that FlptFToA produced.
Maybe there is a better solution, but since I knew I wanted four digits
after the decimal, I did the following:
- multiply the float by 10,000 (moves decimal place 4 places to right)
- typecast the value to a long int
- for each digit of the resulting integer
- pick off last digit using a mod 10 (num %% 10)
- add to string
- after four digits added to string, insert a decimal point into string
- divide number by 10
- reverse the string
I know it is kind of brute force, but it works.
Mike
-----Original Message-----
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED]]On Behalf Of David
Leland
Sent: Thursday, March 01, 2001 10:00 AM
To: Palm Developer Forum
Subject: Re: Determining number of decimal positions
I was afraid that might be the case. So let me ask hypothetically, if the
Palm OS supported the "%f" formatting modifier on the StrPrintF function
(e.g. returnedString = StrPrintF("%f", floatValue);), what value would be
placed into returnedString if the float values were:
12.3450
123.400
1234.56
1234.50
12345.00
Dave
Mike Walters <[EMAIL PROTECTED]> wrote in message news:41279@palm-dev-forum...
>
> By "number of decimal places" do you mean the number before the decimal,
the
> number after, or both?
>
> 1) For the number before the decimal place: I don't know of a built-in
> function, but you could use a small piece of code like the following to
> count the number of places:
>
> int CountDigits (float your_val)
> {
> int count;
>
> count = 0;
> while (your_val > 1.0)
> {
> count ++;
> your_val /= 10.0;
> }
> return count;
> }
>
> 2) For the number of digits after the decimal, the answer really can't be
> determined (do you count all of the trailing zeros, worry arount floating
> point rounding, etc). So the easiest thing to do is just select a value
to
> use in displaying your number, then truncate off any trailing zeros that
you
> don't want.
>
> 3) For the total number of digits, you could probably use a combination of
> the two methods above.
>
> Of course, for any kind of problem, you are going to find almost as many
> solutions as you have responses. An without having the M&R book in front
of
> me, I'm not sure the value that it is expecting to be passed.
>
> Hope this helps,
>
> Mike Walters
> Rose Software
>
>
>
> -----Original Message-----
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED]]On Behalf Of David
> Leland
> Sent: Thursday, March 01, 2001 9:28 AM
> To: Palm Developer Forum
> Subject: Determining number of decimal positions
>
>
> I have a function that is being passed a float value. In my function, I
> will be formatting that float value to a string (e.g. 1234.56 to
1,234.56)
> and returning it. My problem is that I do not know the number of decimal
> places being passed. The float value could be 123.456 or it could be
> 1234.56 or 12.3456. Is there a way I can determine how many decimal
> positions that float value has?
>
> "Advanced Palm Programming" by Mann & Rischpater has an excellent routine
> for formatting a float value to a string but unfortunately, it expects the
> calling function to tell it how many decimal places to format. My App is
a
> Plug-in to another Palm App so I don't have the luxury of knowing upfront
> how many decimal places the float value has.
>
> Thanks,
> Dave
>
>
>
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>
>
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