Round

[grzesiek]

Round(36.285, 2) returns 36.28

Round(36.275, 2) returns also 36.28

Can someone confirm this?

[MikahS]

Local $val1 = Round(36.285, 2)
Local $val2 = Round(36.275, 2)

MsgBox(0, "", "val1: " & $val1 & @CRLF & "val2: " & $val2)

Yes, I get the same result.

[iamtheky]

more fun

msgbox(0, '.275' , Round(number(.275), 2))  ; right
msgbox(0, '.285' , Round(number(.285), 2))  ; wrong

msgbox(0, '1.275' , Round(number(1.275), 2))  ; wrong
msgbox(0, '1.285' , Round(number(1.285), 2))  ; right

msgbox(0, '36.275' , Round(number(36.275), 2))  ;right
msgbox(0, '36.285' , Round(number(36.285), 2))  ;wrong

msgbox(0, '.555' , Round(number(.555), 2)) ;right
msgbox(0, '.565' , Round(number(.565), 2)) ;wrong

Edited April 15, 2015 by boththose

[MikahS]

Something to do with the 2nd decimal spot if it is an 8.

Edited April 15, 2015 by MikahS

[iamtheky]

but it works as expected for every other attempt (of x.275 and x.285), those are the only ones I have been able to break it with so far.  Definitely seems odd, maybe a lesson in floating-point math is upcoming. 

 

***added an example where .555 rounds correctly but .565 does not.

Edited April 15, 2015 by boththose

[MikahS]

Definitely seems odd, maybe a lesson in floating-point math is upcoming. 

Seems this might be the case. 

[Spandex]

FWIW, Excel gives the same results:

=MROUND(36.275,0.01)    results in    36.28
=MROUND(36.285,0.01)    results in    36.28
 

[AspirinJunkie]

I think it's related to the binary rounding error:

ConsoleWrite(StringFormat("%.3f = %.18f\n%.3f = %.18f\n", 0.555, 0.555, 0.565, 0.565))

[Yashied]

The problem is that if the internal representation of the floating point number is 0.2849... instead of 0.285, the rounding algorithm will incorrectly assume the last digit is a 4 and not a 5 and then return the lower number instead of the higher one.

 

ConsoleWrite(StringFormat('%.20f', 0.285) & @CRLF) ;==> 0.28499999999999998000

[iamtheky]

does this work?

msgbox(0, '.275' , _Round(number(.275), 2)) ; right
msgbox(0, '.285' , _Round(number(.285), 2)) ; wrong

msgbox(0, '1.275' , _Round(number(1.275), 2)) ; wrong
msgbox(0, '1.285' , _Round(number(1.285), 2)) ; right

msgbox(0, '36.275' , _Round(number(36.275), 2)) ;right
msgbox(0, '36.285' , _Round(number(36.285), 2)) ;wrong

msgbox(0, '.555' , _Round(number(.555), 2)) ;right
msgbox(0, '.565' , _Round(number(.565), 2)) ;wrong


Func _Round($number , $nPlaces)

$x = round(number($number) , $nPlaces)

If number($x) < number($number) Then
$sPlusOne = "." & stringformat("%0" & $nPlaces & "s" , "1")
$x = number($x) + number($sPlusOne)
EndIf

return $x

EndFunc

[jchd]

Of course this would need to cope with negative values as well, but even then I wouldn't be surprised if counter-examples exist.

BTW I think the help of Number is misleading: if flag = 0 means flag = Default, then we should get an integer which isn't the case.

[SadBunny]

ConsoleWrite(StringFormat("%i", 4.6 * 100) & @crlf)
ConsoleWrite(StringFormat("%0.0f", 4.6 * 100) & @crlf)

You would say this gives the same result twice. But it doesn't:

459
460

So apparently 4.6 * 100 = 459 if you expect an integer (yes, those are some very low values of 4.6 and 100) but 460 if you expect a float.