Re: [ExcelVBA] failing function

 

Dear Greg

This worked for me :-

Public Function Func(x As Range, y As Range) As Variant

' ---- my code ----

    Func = "My result"

End Function

In the spreadsheet I put :-

=PERSONAL.XLSB!Func( A5:E5,A6:E6), -  i.e. the full path to the Function.

You cannot directly run a Function in the VBA IDE, you have to call it from a Sub. That is why it asks for a Macro name.

You don't need the ByVal unless you are modifying the ranges in your code. ByRef is implicit.

Returning a Variant type to the spreadsheet makes it more resilient while you are developing.

Regards

Derek Turner

England

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From: "'David Smart' smartware.consulting@gmail.com [ExcelVBA]" <ExcelVBA@yahoogroups.com>
To: ExcelVBA@yahoogroups.com
Sent: Friday, 21 October 2016, 23:50
Subject: Re: [ExcelVBA] failing function

 



"Public"?  I.e.

 

Public Function Func(ByVal x as range, y as range) as variant

 

Not sure about the ByVal on the range, either.  You might need ByRef on both parameters.  As in

 

Public Function Func(ByRef x as range, ByRef y as range) as variant

 

Lastly, avoid a return type of Variant if you can - give it a more appropriate return type (unless, of course, you are returning various types).

Regards, Dave S
 

----- Original Message -----

From: solutionbuilder@yahoo.com [ExcelVBA]

To: ExcelVBA@yahoogroups.com

Sent: Saturday, October 22, 2016 1:09 AM

Subject: [ExcelVBA] failing function

I am trying to create a Function

I create a module (module1 in General Declaration)

I put in the code

 
Function Func(ByVal x as range, y as range) as variant

---- my code ----

Function End

in the spreadsheet I put in   =Func(A1:E1,A2:E2)

The results are VALUE error

I try to run the Function in VBA Editor and it keeps asking for a Macro Name.

I am missing something simple I am sure, but what is it?

Thanks for responding to this.

 Greg

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Posted by: Derek Turner <g4swy@yahoo.com>

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