e-008 in floats

BlitzMax Forums/BlitzMax Programming/e-008 in floats

Reda Borchardt

(Posted 2009) [#1]

Hello,

I noticed yesterday that BMax sometimes stores floats or doubles with an 'e' notation. How can I convert those into float or doubles that use digits only?

Even casting to a string or a double retains the 'e' notation at the end.

Is there a built in function to do this, or do I need to convert it myself?

Rgds
Reda

Brucey

(Posted 2009) [#2]

stores floats or doubles with an 'e' notation

No. It's only the conversion to a string that makes it appear that way.

Brucey

(Posted 2009) [#3]

If it were me, I'd use this :

SuperStrict

Framework BaH.Format
Import BRL.StandardIO

Local formatter:TFormatter = TFormatter.Create("%.22f")

Local f:Double = 33.0/4433300.44

Print "Default   = " + String.FromDouble(f)
Print "Formatted = " + formatter.DoubleArg(f).Format()


' Default   = 7.4436643444641959e-006
' Formatted = 0.0000074436643444641959

Other people will do their own thing, no doubt.

ziggy

(Posted 2009) [#4]

The 'e' means *10^
As instance:

Local v:Float = 0.00000000009
Print v

will output 9.00000005e-011 wich means: 9 * 10^(-11), or 0.00000000009.

Reda Borchardt

(Posted 2009) [#5]

Brucey,

Thank you very much. That seems to be the solution.

matibee

(Posted 2009) [#6]

Hey Reda,

I think Bruce and Ziggy know more than me, and perhaps you know already about the issues involved in working with floats, but I'll bring it up anyway..

You may be asking too much of floating values if you expect to be able check them for reaching zero, or comparing them against eachother. The usual case is to check them to being close within a tolerance (as defined by your application / units).

SuperStrict 

Function floatCompare:Int ( f1:Float, f2:Float, tol:Float = 0.001:Float )
	Return (f1 + tol >= f2) And (f1 - tol <= f2 )
End Function 

Local f1:Float = 0.00010:Float
Local f2:Float = 0.00008:float

' comparing two floats to different tolerances
Print floatCompare( f1, f2, 0.00001:Float ) 	' returns FALSE
Print floatCompare( f1, f2 ) 	' returns TRUE

' checking if a float "is approximately zero"
Print floatCompare( f2, 0 )	' returns TRUE

Cheers
Matt

Reda Borchardt

(Posted 2009) [#7]

matibee

I have indeed been thinking about this issue but with no elegant solution. Your code example has definitely put me on the right path.