Round a number function problem

[BasicCoder2]

Are there any math brains here that can show me a simple way to round a single (or double) float value to a rounded value like this using FreeBASIC?

declare function ROUND(f as float) r as integer

input output
SINGLE ROUNDED
-3.0 -3
-2.7 -3
-2.4 -2
-2.1 -2
-1.8 -2
-1.5 -2
-1.2 -1
-0.9 -1
-0.6 -1
-0.3 0
0.0 0
+0.3 0
+0.6 +1
+0.9 +1
+1.2 +1
+1.5 +2
+1.8 +2
+2.1 +2
+2.4 +2
+2.7 +3
+3.0 +3

[fxm]

Simply CINT !

[BasicCoder2]

Almost. I notice that 1.5 becomes 1 but -1.5 becomes -2 in the code example below.
So the round up/down rule for .5 isn't symmetrical between positive and negative number in this program.

Code: Select all

screenres 640,480,32
dim as single ii
for i as single = -3 to 3 step 0.3
    print i;tab(20);CInt(i)
    print "----------------------------------------------"
next i

sleep

[MichaelW]

I would have guessed that CINT uses the CRT, but apparently not.

Code: Select all

#include "crt.bi"

dim as single a1(0 to 20) = {-3.0,-2.7,-2.4,-2.1,-1.8,-1.5,-1.2,-0.9,-0.6, _
                             -0.3, 0.0,+0.3,+0.6,+0.9,+1.2,+1.5,+1.8,+2.1, _
                             +2.4,+2.7,+3.0 }
                             
dim as double a2(0 to 20) = {-3.0,-2.7,-2.4,-2.1,-1.8,-1.5,-1.2,-0.9,-0.6, _
                             -0.3, 0.0,+0.3,+0.6,+0.9,+1.2,+1.5,+1.8,+2.1, _
                             +2.4,+2.7,+3.0 }  
                             
for i as integer = 0 to 20
    print a1(i);chr(9);rintf(a1(i));chr(9);rint(a2(i))
next

sleep

Code: Select all

-3      -3      -3
-2.7    -3      -3
-2.4    -2      -2
-2.1    -2      -2
-1.8    -2      -2
-1.5    -2      -2
-1.2    -1      -1
-0.9    -1      -1
-0.6    -1      -1
-0.3    -0      -0
 0       0       0
 0.3     0       0
 0.6     1       1
 0.9     1       1
 1.2     1       1
 1.5     2       2
 1.8     2       2
 2.1     2       2
 2.4     2       2
 2.7     3       3
 3       3       3

[fxm]

Same result with the FB keyword CINT:

Code: Select all

#include "crt.bi"

dim as single a1(0 to 20) = {-3.0,-2.7,-2.4,-2.1,-1.8,-1.5,-1.2,-0.9,-0.6, _
                             -0.3, 0.0,+0.3,+0.6,+0.9,+1.2,+1.5,+1.8,+2.1, _
                             +2.4,+2.7,+3.0 }
                             
dim as double a2(0 to 20) = {-3.0,-2.7,-2.4,-2.1,-1.8,-1.5,-1.2,-0.9,-0.6, _
                             -0.3, 0.0,+0.3,+0.6,+0.9,+1.2,+1.5,+1.8,+2.1, _
                             +2.4,+2.7,+3.0 }  
                             
for i as integer = 0 to 20
    print a1(i);chr(9);rintf(a1(i));chr(9);rint(a2(i));chr(9);cint(a1(i));chr(9);cint(a2(i))
next

sleep

In fact the rounding depends on program context!

[MichaelW]

Same result with the FB keyword CINT
...
In fact the rounding depends on program context!

Yes, that is the result I get, but why would the result vary with the context?

[fxm]

Depending on program calculation context!

The BasicCoder2's code, but just with 'step=0.5':

Code: Select all

screenres 640,480,32

for i as single = -3 to 3 step 0.5
    print i;tab(20);CInt(i)
    print "----------------------------------------------"
next i

sleep

[badidea]

Remove sign, then add sign again, something like this?

Code: Select all

dim as single a
dim as integer i, j

for i = -10 to +10
	a = i * 0.3
	j = int(abs(a)+0.5)*sgn(a)
	print a, j
next

[D.J.Peters]

In case of SINGLE you can cast it as ULONG and clear the sign before call INT().
In case of DOUBLE you can cast it as ULONGINT and clear the sign before call INT().

here are:
function SNGRound(x as single) as integer
function DBLRound(x as double) as longint

Joshy

Code: Select all

union U_SNG
  as single s
  as ulong  u
end union

function SNGRound(x as single) as integer
  dim as U_SNG u=any
  u.s = x + .5
  dim as ulong s = u.u and &H80000000
  if s then  
    u.u = u.u and &H7FFFFFFF
    u.u = int(u.s)
    return &HFFFFFFFF - u.u
  else
    return int(u.s)
  end if
end function

union U_DBL
  as double   d
  as ulongint u
end union

function DBLRound(x as double) as longint
  dim as U_DBL u=any
  u.d = x + .5
  dim as ulongint s = u.u and &H8000000000000000
  if s then  
    u.u = u.u and &H7FFFFFFFFFFFFFFF
    u.u = int(u.d)
    return &HFFFFFFFFFFFFFFFF - u.u
  else
    return int(u.d)
  end if
end function

dim as single  si
dim as double  di
dim as integer j
dim as longint k

for si = -3 to +3 step .25
  di=si
  j = SNGRound(si)
  k = DBLRound(di)
  print si,j,k
next
sleep

[MrSwiss]

Simpler version (no unions):

Code: Select all

Function SNGRound (ByVal s As Single) As Long
	Dim As Long ti
	If s < 0.0 Then
		s  -= .000001		' add/subtract bias
		s  *= -1			' switch sign
		ti  = CLng(s)		' convert
		ti *= -1			' switch sign
		Return ti
	Else
		s  += .000001		' add/subtract bias
		Return CLng(s)		' convert
	EndIf
End Function

Function DBLRound (ByVal d As Double) As LongInt
	Dim As LongInt ti
	If d < 0.0 Then
		d  -= .0000000001	' add/subtract bias
		d  *= -1			' switch sign
		ti  = CLngInt(d)	' convert
		ti *= -1			' switch sign
		Return ti
	Else
		d  += .0000000001	' add/subtract bias
		Return CLngInt(d)	' convert
	EndIf
End Function


dim as single  si
dim as double  di
dim as Long j
'Dim as Integer j
dim as longint k

for si = -3 to +3 step .25
  di=si
  j = SNGRound(si)
  k = DBLRound(di)
  print si,,j,,k
next
sleep

[BasicCoder2]

So much to digest! Thank you for all your inputs.
The difference between Single and Double surprised me as well.
Whereas the Single returns 3 the Double returned 2.999999999999999

Code: Select all

screenres 640,480,32
locate 1,1
for i as single = -3.0 to 3.0 step 0.3
    print i;tab(25);CInt(i)
    print "----------------------------------------------"
next i
sleep
cls
locate 1,1
for i as double = -3.0 to 3.0 step 0.3
    print i;tab(25);CInt(i)
    print "----------------------------------------------"
next i
while inkey<>"":wend
while inkey="":wend

.

[D.J.Peters]

Simpler version (no unions):

@MrSwiss "Simpler" does not mean better
your code used multiplication :-(

[BasicCoder2]

Remove sign, then add sign again, something like this?

I had the same thoughts and the way you did it seems to work but I wanted it as a function.
Something like this although this one looks to me like yours it doesn't work properly.

Code: Select all

screenres 640,480,32

function round(n as single) as integer
    return int(abs(n)+0.5)*sgn(n)
end function

for i as single = -3 to 3 step 0.3
    print i,round(i)
    print "--------------------------------"
next i

sleep

.

[BasicCoder2]

Just to clarify the problem the context is wanting to plot a reflection of pixels around some center point.
The data is in the form of floats but a pixel has an integer value but negative and positive floats are round down instead of away from the center point. Apparently there are four ways of rounding numbers and the dividing point 0.5 can be chosen either up or down but I would at least want it to be consistent for positive and negative numbers.

Code: Select all

screenres 640,480,32
color rgb(0,0,0),rgb(255,255,255):cls

for i as single = -100.0 to 100.0 step 2.3
    if i<0 then color rgb(255,0,0)
    if i>0 then color rgb(0,0,255)
    pset (i+320,240)
    line (320,0)-(320,238),rgb(0,255,0)
    line (320,242)-(320,479),rgb(0,255,0)
next i

sleep

[badidea]

Whereas the Single returns 3 the Double returned 2.999999999999999

I expect that exactly 3 does not exist for single and double floats. the difference between 3 and 2.999999999999999 is maybe a print formatting issue.