Round Function

[atzoref]

Hi,

When I'm doing "Round(4.515 , 2)" it gives 4.51

How can I make it to get 4.52 instead?

Thanks

Edited April 22, 2012 by atzoref

[water]

You could add a very small amount so values of 5 get rounded up.

ConsoleWrite(Round(4.515  + 0.000000000000001, 2) & @LF)  ; returns 4.52

Or you write or own rounding function.

Edit:

For your special case this works too:

ConsoleWrite(Round(4.515*10, 1)/10 & @LF)

Edited April 22, 2012 by water

[water]

Strange thing:

ConsoleWrite(Round(0.515, 2) & @LF)
ConsoleWrite(Round(1.515, 2) & @LF)
ConsoleWrite(Round(2.515, 2) & @LF)
ConsoleWrite(Round(3.515, 2) & @LF)
ConsoleWrite(Round(4.515, 2) & @LF)
ConsoleWrite(Round(5.515, 2) & @LF)
ConsoleWrite(Round(6.515, 2) & @LF)
ConsoleWrite(Round(7.515, 2) & @LF)
ConsoleWrite(Round(8.515, 2) & @LF)
ConsoleWrite(Round(9.515, 2) & @LF)

returns

0.52
1.52
2.52
3.52
4.51
5.52
6.52
7.52
8.52
9.52

Maybe a bug? I will check.

Edited April 22, 2012 by water

[guinness]

That looks correct to me, 9.515 is 9.52 if you round it likewise if you use 9.514 it returns 9.51, which is to be expected.

[water]

Not entirely. As you can see from the example all rounded results end with .52 but 4.515 which returns .51 (which is wrong).

I think this is a problem with converting floating numbers to the internal binary representation.

What I'm not sure if this is a bug with the round function or something you have to live with.

[guinness]

Ah, I missed it as I briefly skimmed through the output, sorry about that. Looks like a bug to me.

[atzoref]

Strange coincidence

I take this specific number for example randomaly from my table, and you show a difference just with this specific number.

[guinness]

Weird, it seems only 4.515 is returning the incorrect value. Am I missing something?

For $i = 0 To 100
    ConsoleWrite(Round(Number($i & ".515"), 2) & @LF)
Next

[water]

Can anyone explain this behaviour or should I open a ticket?

[Melba23]

Hi,

I am fairly certain that this is the well-known "floating point representation" error manifesting itself. In the particular case of 4.515 the value is represented in such a way that the Round goes down and not up.

I would open a ticket, but I am not sure that it is fixable as it is not an AutoIt bug.

M23

[ProgAndy]

Weird, it seems only 4.515 is returning the incorrect value. Am I missing something?

The problem ist the encoding as a double value. 4.515 cannot be represented exactly, it will be stored as 4.51499.

If you want to see the actual values, binary representation etc you can try it here: http://www.h-schmidt.net/FloatConverter/IEEE754.html

[guinness]

The problem ist the encoding as a double value. 4.515 cannot be represented exactly, it will be stored as 4.51499.

If you want to see the actual values, binary representation etc you can try it here: http://www.h-schmidt.net/FloatConverter/IEEE754.html

OK, thanks ProgAndy. Edited August 25, 2012 by guinness

[water]

Thanks for the explanation. So, no need for a bug report.

[Malkey]

It is not just 4.515. This script is suppose to display at the SciTE console the extent of this inaccurate, double value, presentation.

Local $iRoundDown = 0, $iRoundUp = 0
For $h = 0 To 9
    ConsoleWrite(@LF & "============== Check numbers that rounded down to " & $h & " ========================" & @LF)
    For $i = 0 To 99
        For $j = 0 To 9
            $a = StringFormat("%2.2f", Round(Number($i & "." & $j & $h & "5"), 2))
            If StringRight($a, 1) = $h Then
                ConsoleWrite("#" & $a & "#" & @TAB)
                $iRoundDown += 1
            Else
                ConsoleWrite(" " & $a & " " & @TAB)
                $iRoundUp += 1
            EndIf
        Next
        ConsoleWrite(@LF)
    Next
Next
ConsoleWrite(" Round down = " & $iRoundDown & " out of " & $iRoundDown + $iRoundUp & " (" & $iRoundDown * 100 / ($iRoundDown + $iRoundUp) & "%)" & @LF)

Edit: Added to this example which now consistently returns:-

" Round down = 572 out of 10000 (5.72%)"

Edited April 22, 2012 by Malkey

[water]

ProgAndy's explains it quite well. As I assumed it's a problem related to the internal representation of the number and quite sure it's hence not limited to the one example we have found.

So when working with floating point numbers you always have to check the result

[jchd]

Food for thought about rounding: http://en.wikipedia.org/wiki/Rounding

Floating point operations are a complex domain. After 25 years of IEEE work, we still don't know what A + B should yield exactly in all cases.

[MvGulik]

Checking things with StringFormat() is always a good thing when things look odd.

For $i = 0 to 5
ConsoleWrite(StringFormat('%0.'&string($i)&'f', 4.515) & @CRLF)
Next
;; but ...
ConsoleWrite(StringFormat('%0.20f', 4.515) & @CRLF)
#cs
5
4.5
4.51
4.515
4.5150
4.51500
..
4.51499999999999970000
#ce

... now where is my indentation. grrr. ...

Edited April 22, 2012 by MvGulik

[Malkey]

This may be a better example to show the relevance of the number of decimal places in the StringFormat function.

For $i = 0 To 20
    ConsoleWrite(StringFormat('%2d/ %0.' & $i & 'f', $i, 4.515) & @CRLF)
Next
;; but ...
ConsoleWrite(StringFormat('%0.20f', 4.515) & @CRLF)
#cs
0/ 5
 1/ 4.5
 2/ 4.51
 3/ 4.515
 4/ 4.5150
 5/ 4.51500
 6/ 4.515000
 7/ 4.5150000
 8/ 4.51500000
 9/ 4.515000000
10/ 4.5150000000
11/ 4.51500000000
12/ 4.515000000000
13/ 4.5150000000000
14/ 4.51500000000000
15/ 4.515000000000000
16/ 4.5149999999999997
17/ 4.51499999999999970
18/ 4.514999999999999700
19/ 4.5149999999999997000
20/ 4.51499999999999970000
4.51499999999999970000
#ce

Also, I added to my post #14 example to quantify the results.

[jchd]

As a sidenote, the last 7 in the long printout is wrong and should be a 6.

People have hard time getting that FP can only represent a ridiculously small number of real numbers and the fact that all hardware FP is done in binary.

[ProgAndy]

People have hard time getting that FP can only represent a ridiculously small number of real numbers and the fact that all hardware FP is done in binary.

And do not forget that the higher the absolute value the less precision you get.