_IsOddEven

[LurchMan]

Hey everyone -

I was recently working on a program and I needed to figure out if a number was an odd or even number and after I figured out how to determine this I figured I would share it here in case someone else came across this.

Code and example:

$rtn = _IsOddEven (1)
ConsoleWrite($rtn & @CRLF)
$rtn = _IsOddEven (2)
ConsoleWrite($rtn & @CRLF)
$rtn = _IsOddEven (15987)
ConsoleWrite($rtn & @CRLF)
$rtn = _IsOddEven (15896)
ConsoleWrite($rtn & @CRLF)
$rtn = _IsOddEven ("h15896")
ConsoleWrite($rtn & @CRLF)

Func _IsOddEven ($iNum)
    If Not Number ($iNum) Then Return -1
    $iNum = Number (StringRight($iNum, 1))
    For $iTest = 1 To $iNum+5 Step 2
        If $iNum = $iTest Then Return "Odd"
    Next
    For $iTest = 2 To $iNum+5 Step 2
        If $iNum = $iTest Then Return "Even"
    Next
    Return -1
EndFunc

Its fairly simple - Returns either Odd or Even, or -1 if it is not even a number.

(If someone has already done this then just ignore me. I did a quick search and couldn't find anything for it.)

[guinness]

Nice, adding to my Function Folder!

[LurchMan]

Nice, adding to my Function Folder!

Thank you!

[MrCreatoR]

I use this:

#include <Misc.au3> ;only for _Iif() function - to check conditions

$iNumber = 123456789

$iIsEven = _NumberIsEven($iNumber)

If @error Then
    MsgBox(48, "Even check", "<" & $iNumber & "> is not a number.")
Else
    $sMessage = _Iif($iIsEven, "an Even number", "not an Even number (it is Odd)")
    MsgBox(64, "Even check", "The given number <" & $iNumber & "> is " & $sMessage)
EndIf

Func _NumberIsEven($iNumString)
    If Not StringIsDigit($iNumString) Then
        Return SetError(1, 0, -1)
    EndIf
    
    Return Number(Mod(Number($iNumString), 2) = 0)
EndFunc

It's a proper way to check even number.

[Ascend4nt]

Hmm, didn't think this would deserve a function:

BitAND($iVal,1)

[LurchMan]

I honestly don't know what the BitAnd or Mod functions do. I'm still fairly new to programing.

[Mat]

BitAND returns any bits that are in the first number AND the second, hence the name. In this case, the second has only one bit set. This checks if that bit is set in this case. To understand that, you'll need to understand binary, do some reading - As a simple explanation though: since the bits go 1,2,4,8,16 ... the first bit will always be set for an odd number.

Number: 00101010 (42)
BitAnd: 00000001 (1)
Result: 00000000 (0)

Number: 00101011 (43)
BitAnd: 00000001 (1)
Result: 00000000 (1)

Edit: Also remember that it's right to left in terms of bits... So the first bit is in fact the 8th from the left. It may sound stupid, but it works exactly the same way with normal numbers.

Mod on the other hand is basically the remainder like you learnt it in primary school. 9 / 2 is 4 remainder 1, Likewise Mod(9, 2) returns 1.

On computers, using the bitwise functions (AND, NOT, OR, XOR ... ) should be the fastest way.

Edited February 11, 2011 by Mat