how do I express the term if x is integer in VBA language?

Question

how do I express the term if x is integer in VBA language ? I want to write a code that does something if x is integer and does something else if its not with vba excel.

Sub dim()
  Dim x is Variant

  'if x is integer Then 

  'Else:

End Sub 

Algol

Do you mean if x is an Integer (instead of a different type like Decimal or Long) or if x is an integer (a number with no fractional part, such as 3 instead of 3.2)? – Allhallows

Edit the question, select the part that contains the code & click on the toolbar above the textbox on an icon which has 0s and 1s. That will format the code to show as code. And, go ahead & accept the answer, that helped you getting what you want. – Cocaine

@Cocaine There are other answers that can also help @Algol get what they want. – Copulation

Answer 1

If IsNumeric(x) Then 'it will check if x is a number

If you want to check the type, you could use

If TypeName(x) = "Integer" Then

Cocaine

Thanks for the input. I looked up and found it here. – Horvitz

This answer is wrong: IsNumeric("1.2") (or "1,2" depending on your locale) returns True: msdn.microsoft.com/en-us/vba/language-reference-vba/articles/… – Naturally

@Nickolay: What specifically is wrong esp looking at the question? – Cocaine

An integer is a number without a fractional part; IsNumeric only checks if the given value can be converted to a number, it returns True for non-integer numbers. – Naturally

@Nickolay: Does my edit on using TypeName not help? – Cocaine

The second part is valid, it was the first part that prompted my comment... Look, I didn't mean to imply your answer isn't helpful; "wrong" was maybe a bit too harsh (sorry!). I only wanted to leave a quick note that IsNumeric is not the way to check if a value is integer, let's leave it at that? – Naturally

Answer 2

This may suit:

If x = Int(x) Then

Vannoy

This or the same with fix [ if x = fix(x) then ] worked perfectly for my purposes since Int and Fix are equivalent other than for negative non-integers. – Reconstructionism

Answer 3

It depends on if you mean the data type "Integer", or Integer in the sense of: "A number with no decimal." If you meant the latter, then a quick manual test will suffice (see first example); if you meant the former then there are three ways to look at data types all with various pros and cons:

1. VarType will tell you the variant's sub type (see example). It's reasonably fast as it's a just a memory read to an enum, but can only be used on variants and won't tell you the specific object type. Additionally the variant's sub type is often assigned automatically (generally tending to type with the smallest suitable datatype). But it can be tricky (see example).
2. TypeName is the most flexible and robust. It can tell you specific class types and Variant sub-types. It has a slight drawback in that to test requires a string comparison, so there is a minuscule performance hit. It won't be noticeable unless there is some serious repetition though. Also if you have two Objects of the same name in your project (Example Word.Range and Excel.Range), TypeName can't tell the difference (it will return "Range" for both).
3. Finally there is the TypeOf operator (which won't help you here). The TypeOf operator can not test for primitives (example: Integer, Long, String, etc.) but it can tell the specific type of object (example. Excel.Range vs Word.Range). The TypeOf operator will throw an error if the object is nothing, so you must always pretest the object for "Not Is Nothing" but being an operator rather than a function, it is much faster than TypeName (even with the pretest).

Public Sub ExampleManual()
    Dim d As Double
    d = 1
    If Fix(d) = d Then
        MsgBox "Integer"
    End If
End Sub

Public Sub ExampleTypeName()
    Dim x As Integer
    MsgBox TypeName(x)
End Sub

Public Sub ExampleTypeOf()
    Dim x As Excel.Range
    Set x = Selection
    ''//Using TypeOf on Objects set to Nothing will throw an error.
    If Not x Is Nothing Then
        If TypeOf x Is Excel.Range Then
            MsgBox "Range"
        End If
    End If
End Sub

Public Sub ExampleVarType()
    Dim x As Variant
    ''//These are all different types:
    x = "1"
    x = 1
    x = 1&
    x = 1#
    Select Case VarType(x)
        Case vbEmpty
        Case vbNull
        Case vbInteger
        Case vbLong
        Case vbSingle
        Case vbDouble
        Case vbCurrency
        Case vbDate
        Case vbString
        Case vbObject
        Case vbError
        Case vbBoolean
        Case vbVariant
        Case vbDataObject
        Case vbDecimal
        Case vbByte
        Case vbUserDefinedType
        Case vbArray
        Case Else
    End Select
End Sub

Kellykellyann

There is a serious limitation with TypeName. Type "3" in cell A1 and look at the output of TypeName(Range("A1").value). On my system I get "Double". By default numbers on a Worksheet seem to have a Double type, even when they don't have a fractional part. – Cyanotype

Answer 4

None of the answers worked for me. If the input is anything EG a String then this code is safe to run without errors:

Function IsInt(s As String) As Boolean
    Dim res As Boolean: res = False
    If IsNumeric(s) Then
        res = (s = Int(s))
    End If
    IsInt = res
End Function

Note: apparently VBA doesn't have short-circuit evaluation (SCE) so you need an if-then statement here.

I don't appreciate the absence of SCE, but at least now I have an example to show to my friends and colleagues who assume that all languages have it, while there are arguments both for and against it.

Mash

Answer 5

I was getting (seemingly) random vartypes from user input on userforms. What was indeed an integer may have been returned as a range, string, or other dataype. I've found this method to work best for me when trying to determine if a user input is indeed an integer. It will first check if X is numeric, then check if it's an integer and adds it to a long variable if it is:

Dim x As Variant, i As Long
x = "1"

If IsNumeric(x) Then
    Debug.Print "x is a number"
    If Val(x) = Int(x) Then
        Debug.Print "x IS an integer"
        i = CLng(x)
    Else
        Debug.Print "x is NOT an integer"
    End If
Else
    Debug.Print "x is not a number"
End If

I like the idea of a simple one liner, but the following will not show that X is an integer:

Dim x As Variant
x = "1"
If x = Int(x) Then Debug.Print "X is an integer"

This also did not show that X is an integer:

Dim x As Variant
x = "1"
If IsNumeric(x) Then
    Debug.Print "x is numeric"
    If TypeName(x) = "Integer" Then Debug.Print "x is an integer"
Else
    Debug.Print "x is not numeric"
End If

Follicle