Why does a python COM server return different data to VBA and Python clients?

[jeffwelch]

My apologies to the list about my earlier question that was in html. My webmail is supposed to default to plain text but clearly did not.

I am trying to write a Python server that will be used by a C++ client. It has taken me quite a while longer than I expected. I have been through the archives and that helped me get to where I am. Many thanks to all those who support these packages.

I am using Python 2.5.1 as it is required by some vendor equipment that I must use.

One of the reasons it took me a while to get this going was that I was trying to test it in Python. Once I switched to testing it in VBA then I made much faster progress.

For demonstration purposes I used the Pippo test in the 'win32com\test' directory. I modified the testpippo.py by adding some code at the start of the testLeaks method. I added the 4 lines:

        in1 = 15
        inout1 = 99
        hr= self.object.Method3( in1, inout1)
        print hr, "retval = %d inout1 = %d"%(hr[0], hr[1])


The printed line was "(0, -31) retval = 0 inout1 = -31".


I get "retval = -41 inout1= -31"  if I execute the VBA code:

Sub TestPippo()
    Dim obj As Pippo
    Dim in1 As Long
    Dim inout1 As Long

    Set obj = CreateObject("Python.Test.Pippo")
    in1 = 1
    inout1 = 2
    retval = obj.Method3(in1, inout1)
    MsgBox ("retval = " & retval & " inout1 = " & inout1)
   
End Sub


Is there a way to call the same Python Server Method, Method3, so that both Python and VBA receive the same return data?

Regards,
j
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[Mark Hammond-4]

On 10/04/2012 3:40 AM, [hidden email] wrote:

> My apologies to the list about my earlier question that was in html. My webmail is supposed to default to plain text but clearly did not.
>
> I am trying to write a Python server that will be used by a C++ client. It has taken me quite a while longer than I expected. I have been through the archives and that helped me get to where I am. Many thanks to all those who support these packages.
>
> I am using Python 2.5.1 as it is required by some vendor equipment that I must use.
>
> One of the reasons it took me a while to get this going was that I was trying to test it in Python. Once I switched to testing it in VBA then I made much faster progress.
>
> For demonstration purposes I used the Pippo test in the 'win32com\test' directory. I modified the testpippo.py by adding some code at the start of the testLeaks method. I added the 4 lines:
>
>          in1 = 15
>          inout1 = 99
>          hr= self.object.Method3( in1, inout1)
>          print hr, "retval = %d inout1 = %d"%(hr[0], hr[1])

 >
 >
 > The printed line was "(0, -31) retval = 0 inout1 = -31".
 >

what is Method3 here?  The pippo tests don't have such a method.

But note that in Python, you generally don't see the "hresult" - so it
would be more accurate to have it read something like:

          results = self.object.Method3( in1, inout1)

as Python doesn't have a concept of "byref" args, the "results" object
is a tuple with the return value of the function, if any, plus any
"byref" args created.  In the example above, I'm guessing the function
has returned 0 and one byref arg was presented, with the new value being
-31.

In the VB example below, VB does have byrefs, so the out args end up in
the same argument they were passed in.

I can't explain why VB for -41 for the result when Python got zero
though, but I also can't find the "method3" to help explain it.

Mark

>
> I get "retval = -41 inout1= -31"  if I execute the VBA code:
>
> Sub TestPippo()
>      Dim obj As Pippo
>      Dim in1 As Long
>      Dim inout1 As Long
>
>      Set obj = CreateObject("Python.Test.Pippo")
>      in1 = 1
>      inout1 = 2
>      retval = obj.Method3(in1, inout1)
>      MsgBox ("retval = " & retval & " inout1 = " & inout1)
>
> End Sub
>
>
> Is there a way to call the same Python Server Method, Method3, so that both Python and VBA receive the same return data?
>
> Regards,
> j
> _______________________________________________
> python-win32 mailing list
> [hidden email]
> http://mail.python.org/mailman/listinfo/python-win32
>

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[dived38295]

Thanks for the reply. I forgot that I had copied Method2 and Method3 from a thread I saw from 2006, 2006-October/005094. That thread just seemed to die without resolution.

I was going to include the differences but I determined why it failed. According to the Microsoft support website "By default, VBA passes arguments by reference."

I would have expected VBA to make an "[in]" parameter use ByVal but that didn't happen, so I did some digging. I found that my example was returning more data than expected. If I removed the extra return value it works and the "[in]" parameter from VBA is not overwritten.
If I forward declare the method in VBA then it detects the extra data and reports an error. I find the VBA behavior a bit confusing.

For anyone interested you can see this behavior using the Pippo test with the following changes:

I added the following line to the "pippo.idl" interface IPippo:
        [id(4), helpstring("method Method3")] HRESULT Method3([in] long in1, [in, out] long *inout1, [out, retval] long *val);
        [id(99), helpstring("method Method4")] HRESULT Method4([in] long in1, [in, out] long *inout1, [out, retval] long *val);

I added the following lines to the pippo_server.py class CPippo. Note that Method 4 is returning an extra value.
    def Method3(self, in1, inout1):
        return( 1010,1011 )
    def Method4(self, in1, inout1):
        return( 1010,1011,1012 )
               
Execute the following VBA code (Excel, or Word have VBA in macros):
        Private Declare Function Pippo_Method3 Lib "TESTSERVERLib.Pippo" (ByVal in1 As Long, inout1 As Long) As Long
        Sub TestPippo()
                  Dim obj As Object
                  Dim in1 As Long
                  Dim inout1 As Long

                  Set obj = CreateObject("Python.Test.Pippo")
                 
                  in1 = -1
                  inout1 = -2
                  retval3 = obj.Method3(in1, inout1)
                  outputMessage = outputMessage & "    Forward declared - Method 3 retval3 = " & retval3 & " inout1 = " & inout1 & " in1 = " & in1 & vbCrLf

                  in1 = -1
                  inout1 = -2
                  retval4 = obj.Method4(in1, inout1)
                  outputMessage = outputMessage & "Not Forward declared - Method 4 retval4 = " & retval4 & " inout1 = " & inout1 & " in1 = " & in1 & vbCrLf

                  MsgBox outputMessage

 End Sub

If you call Method4 from a Python client then it only sees a tuple with 2 elements as expected by the idl. It did seem odd that the VBA saw more data so I guess the Python server method return data is not checked against the registered tlb and just passes the data directly through to the COM client.

J
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