DOCUMENT:Q143082  03-AUG-2001  [visualc]
TITLE   :FIX: Wrong Pointer Value When Nested Classes Have Same Name
PRODUCT :Microsoft C Compiler
PROD/VER:
OPER/SYS:
KEYWORDS:kbnokeyword kbVC200bug kbVC210bug kbVC220bug kbVC400bug kbVC410bug kbVC420bug kbVC500bu

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The information in this article applies to:

 - The C/C++ Compiler (CL.EXE), used with:
    - Microsoft Visual C++, 32-bit Editions, versions 2.0, 2.1, 2.2, 4.0, 4.1 
    - Microsoft Visual C++, 32-bit Enterprise Edition, version 4.2 
    - Microsoft Visual C++, 32-bit Professional Edition, version 4.2 
    - Microsoft Visual C++, 32-bit Enterprise Edition, version 5.0 
    - Microsoft Visual C++, 32-bit Professional Edition, version 5.0 
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SYMPTOMS
========

When a class is derived from multiple classes such that two or more of the base
classes are nested classes of the same name, pointers to either of the nested
base classes point to the same address. Consider a class D which is derived from
both B1::Nested and B2::Nested. Given an object d, which is of type D, then
(B1::Nested *)&d and (B2::Nested *)&d will both resolve to the same
address. If either of the nested base class' names are changed to be unique, the
behavior is normal.

RESOLUTION
==========

This happens only when the nested classes have the same name. Change the names,
for example "A::NestedA" and "B::NestedB".

STATUS
======

This bug was corrected in Microsoft Visual C++, version 6.0.

MORE INFORMATION
================

Sample Code to Reproduce Error
------------------------------

   // Compile option needed: none

   // File test.cpp

   truct A {

      struct Nested {
          virtual void A_Func() = 0;
      };

   };

   truct B {

      struct Nested {
          virtual void B_Func() = 0;
      };

   };

   truct MyClass : public A::Nested, public B::Nested
   {

      void A_Func() { cout << "A_Func() called" << endl; }
      void B_Func() { cout << "B_Func() called" << endl; }
   };

   void main()
   {

      MyClass m;
      cout << "(A::Nested*)&m = " << (void*)(A::Nested*)&m << endl;
      cout << "(B::Nested*)&m = " << (void*)(B::Nested*)&m << endl;
      ((A::Nested*)&m)->A_Func();
      ((B::Nested*)&m)->B_Func();

   }

In this example, the compiler confuses (A::Nested*) and (B::Nested*), so the same
values are printed for both ((A::Nested*)&m) and ((B::Nested*)&m) when
they should be different. This also causes the call to B_Func on the last line
to call A_Func instead.

Additional query words:

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Keywords          : kbnokeyword kbVC200bug kbVC210bug kbVC220bug kbVC400bug kbVC410bug kbVC420bug kbVC500bug kbVC600fix 
Technology        : kbVCsearch kbAudDeveloper kbCVCComp
Issue type        : kbbug
Solution Type     : kbfix

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