Round() Command missing?

Am i blind? I cannot find any command that can round Floats to a special accuracy like:

float#=1.23523452

newfloat#=Round(float#,3)

newfloat# will now be 1.235

I know, i could write a function which will do the job - but i am not willing to always code-around missing basic commands. i think such a command is essential and should be integrated by mark. or am i wrong?

I can't think of a highly practical uses for it

You could always do:

num$="1.2345"

Print Left(num$,4)

this would print "1.23" on the screen

yeah but it does not round it up correctly. i have lots of practical uses for it (especially for my game).

Play with this:

Function Round#(Value#,Digits)
 Local Newval#
 if digits>1 then
  newval#=int(value#*(10^(digits-1)))/(10^(digits-1))
  else
  newval#=int(value#)
 endif
 Return newval
End Function

[Edit]

Untested!

Function Round#(val#,precision)
  val#=Floor#(val# * 10^precision)
  val#=val/10^precision
  return val#
end function

Same code using Int to round instead of truncate:

Function Round#(val#,precision)
If precision=0 Then 
	Return Int(val#)
Else
  	val#=Int(val# * 10^precision)
 	val#=val/10^precision
  	Return val#
End If
End Function

kevin, your code does not round correctly.

cygnus, thanks! (though you have a little typo in your code)

true, my code truncates at whatever precision you ask for.

@Cygnus: Noticed that too Val to value#

Still, think it'll work though

Hi, I agree this should be built in to blitz.
I'll add this to the code archives.

'Examples:
Local a:Float = 11.62552
Print "Rounding "+a
Print round(a,5) 'nearest mutiple of 5
Print round(a,2) 'nearest even number
Print round(a,1) 'nearest int
Print round(a,.1) 'one decimal place
Print round(a,.001) 'three decimal places


Function Round:Float(num:Float,precision:Float)
	Return Int(num/precision+.5*Sgn(num))*precision
EndFunction

yeah, Val should have been Val#.

I typed it straight into the post window, as i said, it would need playing with :)

Kevin- your code is functionaly the same as mine, but does it work when you round to one place? (to me, one place means the INT value, two places is INT plus one of the following floats. You also need to use INT instead of FLOOR :) That should do what he wants.

[edit]

I updated the code.

it works from precision of 0 on upwards...truncates, though.

@Cygnus. Shouldn't it be..

Function Round#(value#,Digits)

Surely it would have to return a string? Floating point numbers aren't stored to x digits are they?

Hahaha! Yes thats right :) I fixed it, must not have posted the change!

Kevin, thats cool. Your code should be more efficient than mine then. I programmed my code to accept 1 to be the integer part and 2 to be after the floating point etc.

Still, you should use INT instead of FLOOR to get it to round correctly.

Floor and Ceil commands round up or down.

just a small alteration
this returns a value rounded to the number of decimal
places entered

Function Round#(Value#,Digits)
Local Newval#
If digits>1 Then
newval#=Int(value#*(10^(digits))+0.5)/(10^(digits))
Else
newval#=Int(value#)
EndIf
Return newval
End Function

This is the one I use, and after test, it seams to be the fastest way.
=> (*) is faster than (/)
and x*x is faster than x^n

so, it's optimised to run fast with lot of uses.

fl#=1.2345678901
For n=0 To 10
	Print LIB_GET_Decimal(fl,n)
Next
WaitKey
End


Function LIB_GET_Decimal#(Var#,Dec%=0)
	Local l_D#,l_f#
	Select Dec
		Case 0
			l_D=1
			l_f=1
		Case 1
			l_D=10
			l_f=.1
		Case 2
			l_D=100
			l_f=.01
		Case 3
			l_D=1000
			l_f=.001
		Case 4
			l_D=10000
			l_f=.0001
		Case 5
			l_D=100000
			l_f=.00001
		Case 6
			l_D=1000000
			l_f=.000001
		Case 7
			l_D=10000000
			l_f=.0000001
		Case 8
			l_D=100000000
			l_f=.00000001
		Case 9
			l_D=1000000000
			l_f=.000000001
		Default
			Return Var
	End Select
	v2%	=	Var*l_D
	Var	=	Float(v2)*l_f
	Return Var
End Function

Remeber that Blitz3D don't remember more than 6 numbers after comma.