**Please Note: **
This article is written for users of the following Microsoft Excel versions: 2007 and 2010. If you are using an earlier version (Excel 2003 or earlier), *this tip may not work for you*. For a version of this tip written specifically for earlier versions of Excel, click here: Determining If a Number is Odd or Even.

A common programming task is validating user input. Often, your macro may need to determine if a number entered by a user is odd or even. For instance, suppose you wrote your own macro that asked the user what worksheet number they wanted to process. If your macro had to process odd and even worksheets differently, then you need to figure out if the number the user provided was odd or even. The technique for this is relatively simple, as shown here:

Even = (UserNum Mod 2) - 1

After execution of this line, Even will be True (-1) if UserNum was even, or False (0) if UserNum was odd.

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This tip (11398) applies to Microsoft Excel 2007 and 2010. You can find a version of this tip for the older menu interface of Excel here: **Determining If a Number is Odd or Even**.

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2015-05-26 14:11:53

Aldo

OOPS ... made a typo ...

It should be this:

A / 2 = Int(A / 2) ... in 1.83203125 sec

2015-05-26 14:09:45

Aldo

A = Int(A / 2) ... in 1.615234375 sec

(A Mod 2) = 0 ... in 1.0390625 sec

(A Mod 2) - 1 ... in 1.2734375 sec

Application.IsOdd(A) ... in 275.92578125 sec

WorksheetFunction.IsEven(A) ... in 96.341796875 sec

These time were generated on a 100,000,000 For...Next loop.

Clearly (A Mod 2) = 0 is the fastest; over 20% faster than the method shown here. At the end of the day, most macro/VBA "normal" use wouldn't really notice any difference in performance using any even/odd check method above. But if you are a programmer/developer who wants to optimize your code, consider native commands as often as possible in VBA.

2015-05-25 05:46:36

Well, I stand corrected.

2015-05-24 23:32:44

Locke Garmin

(i Mod 2) - 1

1.70 Seconds

(i Mod 2) = 0

1.52 Seconds

Hardly significant savings but interesting! :)

2015-05-24 23:22:17

LockeGarmin

I got interested in the idea of the worksheet functions being slower that I did a couple of tests. I ran a for loop 1,000,000 on each method like the following:

Dim b As Boolean

Dim i As Long: i = 2

dim x as Long

'Start Stopwatch Macro

For x = 1 To 1000000

b = (i Mod 2) - 1

Next x

'End Stopwatch Macro

And came to the following results:

(i Mod 2) - 1

0.02 Seconds

Excel.WorksheetFunction.IsEven(i)

2.19 Seconds

Application.WorksheetFunction.IsEven(i)

2.27 Seconds

Application.IsEven(i)

5.80 Seconds

So I concluded that there is a small time penalty you would start running into if you start testing quite a few numbers for being even or odd. I hope there is another VBA nut who finds this as interesting as I did! :)

2015-05-23 05:49:36

I don't think you get a time penalty for using a WorksheetFunction, it only happens if you go get a value from the book. Which is once in both examples.

Furthermore, native functions tend to be faster than custom made.

2015-05-22 19:03:10

Michael (Micky) Avidan

To my opinion the average Excelist isn't familiar with the results -1 and 0.

He wants to get a straight answer like TRUE or FALSE.

Therefore my suggestion would be one of the following (Short) commands:

'-------------------

Sub Is_Odd()

MsgBox Application.IsOdd([A1])

MsgBox CBool([A1] Mod 2)

End Sub

'--------

Michael (Micky) Avidan

“Microsoft® Answers" - Wiki author & Forums Moderator

“Microsoft®” MVP – Excel (2009-2015)

ISRAEL

2015-05-22 12:26:29

awyatt

x = Application.WorksheetFunction.ISEVEN(Range("A1").Value)

x = (Range("A1").Value Mod 2) - 1

As a matter of preference, the second is better. It is not only shorter, but it doesn't have the time penalty of using the WorksheetFunction method. The MOD operator doesn't make it it "so complicated" (as Eric stated); it actually makes it less complicated and faster.

-Allen

2015-05-22 11:03:14

Eric Augusta

Dim x as Boolean

x = Application.WorksheetFunction.ISEVEN(Range("A1").Value)

2015-05-22 10:22:26

awyatt

And, as is stated in the first two sentence of this tip, this is for use in a macro.

-Allen

2015-05-22 10:20:26

icy322

What's wrong with using the iseven and isodd funtion?=iseven(a1) or = isodd(a1)

2015-05-22 10:16:42

awyatt

-Allen

2015-05-22 10:05:08

Gary Lundblad

Thank you!

Gary Lundblad

2015-05-22 07:30:18

In that case the workaround would be to test the last digit of the number. Something like:

Val(Right(X,1)) Mod 2

...Supposing the number is an integer. ANd seeing how CInt and CLng don't work with big numbers, you are stuck.

Two solutions:

- Treat the number as text. Remember that Excel has a native limitation of 15 digits per number, so if your number is bigger than 1e15 it won't work.

- Use an addon specialized in big numbers. My favourite is xNum:http://www.bowdoin.edu/~rdelevie/excellaneous/

2012-02-25 14:38:43

Ken Kast

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