Issue with math: float calculation, ToStr()

FUNCTION checkStrFloat(Str AS STRING) AS STRING ' trig by getItemDscr(1), c2Inventory(1)	' Search for a . in the string, then check if the string is longer than one character past the . and if so, trim off the rest
	StrLen = Str.Len()
	FOR i = 1 TO StrLen
		chr = Str.Mid(i -1, 1)								' String.Mid() is 0 based, not 1 based, so need to -1
		IF chr = "."																				' Ex: "9.900001", i will be 2
			IF StrLen > i +1																		' If there is more than just one character after the decimal point
				StrNEW = Left(Str, i+1)						' Left() is 1 based		' Ex: i +1 will be 3, thus first three characters of "9.900001" will be "9.9", one character past the "."
				? "checkStrFloat() found long float Str, orig " Str ", shortened " StrNew
				RETURN StrNEW
			END IF
		END IF
	END FOR
	RETURN Str
END FUNCTION

function convert(f as double) as string
  a = CInt(f * 10)
  n = a \ 10
  m = abs(a) mod 10
  return n.toStr() + "." + m.toStr()
end function

"Komag" wrote:
Try this:
? 1.1* 9
console prints 9.9

But try this:
? str(1.1 * 9)
console prints 9.900001

x=1.1 * 9

' 1) use Fix to try to force 1 digit of precision.  may or may not work?
? Str(Fix(x * 10)/10)

' 2) use printf-style conversion
? x.ToStr("%3.1f")

' Note that Str prefixes a blank on non-negative numbers, the above ToStr does not.
' if you want that, use x.ToStr("% 3.1f")

"renojim" wrote:
My advice is to only use integer math unless there's a compelling reason not to (which there never is).  

? str(1.1*9)
 9.900001
? (1.1*9).tostr()
9.9

FUNCTION washFloat(dirtyFloat)
 Int10 = Cint(dirtyFloat *10)
 cleanFloat = Int10 / 10
 RETURN cleanFloat
END FUNCTION

"Komag" wrote:

And then I'll use float.ToStr() for double measure (unless there are Roku models or firmware on which that won't work?)