Stripping extra decimal spaces

BlitzMax Forums/BlitzMax Beginners Area/Stripping extra decimal spaces

SkyCube

(Posted 2010) [#1]

Hi all

I made a simple salary calculator as a test to learn a bit of MaxGUI. This is a pretty basic question but I can't find an answer in the docs. How do I round a Long number to, say, two decimal places. The thing is, when I do my calculation, the result comes out like this:

206.0000000000000

Any info will be greatly appreciated!

Czar Flavius

(Posted 2010) [#2]

You can convert it to a string, and then remove all characters that occur N characters after the "."

Evil Roy Ferguson

(Posted 2010) [#3]

Function RoundedToDecimalPlaces:String(number:Double, places:Int)
	Local numberString:String = String(number)
	Return numberString[0..numberString.Find(".")+places+1]
End Function

Brucey

(Posted 2010) [#4]

Assuming it has "." in there, otherwise it's going to return something entirely unexpected.

GfK	(Posted 2010) [#5]

Assuming it has "." in there, otherwise it's going to return something entirely unexpected

A double will always have a decimal point in it, won't it?

Evil Roy Ferguson

(Posted 2010) [#6]

At least on Mac and Windows, using the current BMX, BlitzMax's String representation of a Double always contains a '.' as far as I can tell. I couldn't find a counterexample, anyway.

It is the sort of thing that could get changed between updates, though.

Brucey

(Posted 2010) [#7]

A double will always have a decimal point in it, won't it?

Let's hope so.

Jesse

(Posted 2010) [#8]

you can end up with a scientific notation valid on intel and ppc computers.
try this:

 a! = 20.0!/21808999899!
Print a!

GfK	(Posted 2010) [#9]

Oh yeah.... you could parse the string to allow for that though.

Evil Roy Ferguson

(Posted 2010) [#10]

I would like to see a salary involving scientific notation. :)

That having been said, I definitely stand corrected.

THAT having been said, when dealing with monetary values, it's best to avoid the whole floating-point thing anyway -- you might want to consider either using arbitrary-precision arithmetic or storing monetary values as a number of cents (ie, $7.82 as 782) in order to avoid this sort of problem.

Warpy

(Posted 2010) [#11]

Something like this?

Function decround$(n!,k:Int)
	Local i:Int
	i=Int(n)
	If Floor(n)=n Or k=0 Return String(i)
	n:-i
	n:*10^k
	Local d:Int=Int(n)
	If n-d>.5 Then d:+1
	Return String(i)+"."+String(d)
End Function

For k=0 To 5
	Print decround(Pi,k)
Next

Warpy

(Posted 2010) [#12]

Though yes, for salaries store money as numbers of pennies and use bankers' rounding