FIX: _variant_t::operator IUnknown*() Fails to AddRef Interface

• The C/C++ Compiler (CL.EXE) included with:

      - Microsoft Visual C++, 32-bit Editions, version 5.0
  

SYMPTOMS

When using _variant_t::operator IUnknown*(), the returned interface is not appropriately AddRef'd. This can cause unpredictable results when using the returned interface. For instance, subsequent calls through the interface may inexplicably fail or may cause an Access Violation.

RESOLUTION

Write a function to convert a _variant_t to an IUnknown*. A method based on _variant_t::operator IDispatch*() is shown in the sample below.

STATUS

Microsoft has confirmed this to be a bug in the Microsoft products listed at the beginning of this article. This bug has been fixed in Visual Studio 97 Service Pack 3.

For more information, please see the following article in the Microsoft Knowledge Base:

   ARTICLE-ID: Q170365
   TITLE     : INFO: Visual Studio 97 Service Packs - What, Where, and Why

MORE INFORMATION

The code below should be called instead of allowing _variant_t::operator IUnknown*() to be implicitly called to work around this problem. The conversion operator will be called whenever a conversion from _variant_t to IUnknown* is required. One place this occurs is with assignment statements, another is passing a _variant_t to a function that expects an IUnknown*.

Sample

  IUnknown* f( const _variant_t & vt )
  {
      if (V_VT(&vt) == VT_UNKNOWN)
      {
          V_UNKNOWN(&vt)->AddRef();
          return V_UNKNOWN(&vt);
      }
      _variant_t vttmp;

      vttmp.ChangeType(VT_UNKNOWN, &vt);

      V_UNKNOWN(&vttmp)->AddRef();
      return V_UNKNOWN(&vttmp);
   }
   void g()
   {
     _variant_t vt;
     IUnknown * punk=0;
     // ...some code...
     punk = vt;    // will cause _variant_t::operator IUnknown*()
                   // to be called. Comment this out.
     //punk = f(vt); // call f instead. Uncomment for workaround
     // ...more code...
   }